- Identify where both marked angles lie relative to the two parallel lines. - Compare which side of the transversal each angle occupies. - Use the theorem that applies to that angle pair when the lines are parallel.

1. The marked angles lie between the parallel lines on opposite sides of the transversal. 2. They are alternate interior angles, so the Alternate Interior Angles Theorem shows that \(\alpha\cong\beta\).

The Alternate Interior Angles Theorem.

- Identify which fact is given and which fact the proof must establish. - A theorem and its converse reverse the hypothesis and conclusion. - Choose the direction that begins with parallel lines.

1. The parallel relationship \(g\parallel h\) is given. 2. The Corresponding Angles Theorem starts with parallel lines and concludes that corresponding angles are congruent. 3. Its converse works in the opposite direction, starting with congruent corresponding angles to prove that two lines are parallel.

Use the Corresponding Angles Theorem because the parallel lines are given and the conclusion is that the corresponding angles are congruent.

- What is given about the two lines? - Which theorem compares angles in corresponding positions when the lines are parallel?

1. The lines are parallel, and the angles occupy corresponding positions. 2. Corresponding angles formed by a transversal of parallel lines are congruent, so \(\alpha=47^\circ\).

\(\alpha=47^\circ\), because corresponding angles of parallel lines are congruent.

- Are all the angle-pair rules based on equal measures? - Think about the condition for same-side interior angles in particular.

1. Equality is not the condition for same-side interior angles to establish parallelism; these angles must be supplementary. 2. Here their sum is \(80^\circ+80^\circ=160^\circ\), not \(180^\circ\). 3. Leo's reasoning is incorrect. The lines are not parallel.

No. Same-side interior angles must add to \(180^\circ\) to prove lines parallel. These add to \(160^\circ\), so the lines are not parallel.

- Each given sum involves same-side interior angles along one side of the quadrilateral. - What does the converse theorem say when those angles are supplementary?

1. Since \(m\angle E + m\angle F = 180^\circ\), the converse of the same-side interior angles theorem gives \(EH \parallel FG\). 2. Since \(m\angle E + m\angle H = 180^\circ\), the same theorem gives \(EF \parallel HG\). 3. Both pairs of opposite sides are parallel, so \(EFGH\) is a parallelogram.

\(EFGH\) is a parallelogram because \(EH \parallel FG\) and \(EF \parallel HG\).

- Which side connects the vertices of the given angles? - Add the measures of those angles. - What converse uses a supplementary same-side interior pair?

1. Line \(AD\) is a transversal of lines \(AB\) and \(CD\). 2. The two same-side interior angles total \(115^\circ+65^\circ=180^\circ\). 3. Their supplementary sum proves \(AB\parallel CD\) by the converse theorem.

The same-side interior angles add to \(180^\circ\), so \(AB\parallel CD\).

- Name the opposite side pairs parallel in the outer parallelogram. - Which of those lines contain sides of \(ABCD\)? - Combine that fact with the given parallel pair.

1. Because \(NBFD\) is a parallelogram, \(ND\parallel BF\). 2. Segment \(AD\) lies on \(ND\), and \(BC\) lies on \(BF\). Therefore, \(AD\parallel BC\). 3. The given \(AB\parallel CD\) supplies the other pair. 4. Thus \(ABCD\) has both pairs of opposite sides parallel and is a parallelogram.

The outer parallelogram gives \(AD\parallel BC\), and \(AB\parallel CD\) is given, so \(ABCD\) is a parallelogram.

- Identify the relationship between the two angles and the transversal. - Compare their sum with a straight angle. - Use the appropriate converse theorem to infer a relationship between the two lines.

1. The two same-side interior angles have sum \(68^\circ+112^\circ=180^\circ\). 2. If same-side interior angles formed by a transversal are supplementary, then the two lines are parallel. 3. Therefore, \(\ell\parallel n\).

Since \(68^\circ+112^\circ=180^\circ\), the same-side interior angles are supplementary. By the converse of the same-side interior angles theorem, \(\ell\parallel n\).

- Compare the directions of the two perpendicular segments. - What shape has opposite sides along the two given parallel lines? - Which equal-length property does that shape have?

1. Because \(AB\perp\ell\) and \(CD\perp\ell\), these segments are parallel. 2. Points \(A,C\) lie on \(\ell\), and \(B,D\) lie on \(m\), so \(AC\parallel BD\). 3. Quadrilateral \(ACDB\) is therefore a parallelogram with a right angle, hence a rectangle. 4. Opposite sides of a rectangle are congruent, so \(AB=CD\).

The segments form opposite sides of rectangle \(ACDB\), so \(AB=CD\).

- What equal angles come from the bisector \(AD\)? - Transfer one angle using \(DE\parallel AC\). - Which equal angles in triangle \(AED\) give equal opposite sides?

1. The bisector gives \(\angle BAD=\angle DAC\). 2. Because \(DE\parallel AC\), alternate interior angles give \(\angle ADE=\angle DAC\). 3. Since \(E\) lies on \(AB\), \(\angle EAD=\angle BAD\). 4. Therefore, \(\angle EAD=\angle ADE\), and triangle \(AED\) has equal opposite sides \(DE=AE\).

The angle bisector and parallel line give \(\angle EAD=\angle ADE\); thus \(AE=DE\).

- Identify a base shared by all three triangles. - Compare the perpendicular distances from \(C\), \(D\), and \(E\) to that base. - Use the triangle area formula.

1. The three triangles share the same base \(\overline{AB}\). 2. Because \(C\), \(D\), and \(E\) lie on a line parallel to \(AB\), their perpendicular distances to line \(AB\) are equal. 3. Each triangle therefore has the same base length and the same height. 4. Using \(A=\frac{1}{2}bh\), \([ABC]=[ABD]=[ABE]\).

All three triangles have base \(AB\), and their third vertices lie on one line parallel to \(AB\), so their heights are equal. Therefore, their areas are equal.

- How do same-side interior angles establish parallelism? - A vertical angle keeps its measure. What happens when you form a linear pair? - Which two exterior positions are across the transversal from one another?

1. Since \(118^\circ+62^\circ=180^\circ\), the same-side interior angles are supplementary. Their converse theorem gives \(p\parallel q\). 2. The exterior angle vertical to the \(118^\circ\) interior angle also measures \(118^\circ\). 3. The exterior angle forming a linear pair with the \(62^\circ\) interior angle measures \(180^\circ-62^\circ=118^\circ\). These exterior angles are on opposite sides of \(s\), so they are alternate exterior angles. 4. Their equality independently proves \(p\parallel q\) by the alternate exterior angles converse.

The same-side interior angles total \(180^\circ\), so their converse gives \(p\parallel q\). The angle vertical to the \(118^\circ\) angle is \(118^\circ\); the angle adjacent to the \(62^\circ\) angle is also \(118^\circ\). These are alternate exterior angles, and their converse independently gives \(p\parallel q\).

- Treat each side of the triangle as a transversal of the two parallel lines. - Which angles at \(A\) match the triangle's base angles? - What total do the three neighboring angles at \(A\) make?

1. With \(AB\) as a transversal, the angle between \(AB\) and the auxiliary parallel at \(A\) equals \(\angle B=48^\circ\) by alternate interior angles. 2. With \(AC\) as a transversal, the alternate interior angle at \(A\) equals \(\angle C=72^\circ\). 3. The two transferred angles and \(\angle A\) make a straight angle. Thus \(\angle A=180^\circ-48^\circ-72^\circ=60^\circ\).

The two angles are \(48^\circ\) and \(72^\circ\) by alternate interior angles. Therefore, \(\angle A=60^\circ\).

- Which sides of triangle \(ABC\) can also act as transversals? - What angle relationships hold when a transversal cuts parallel lines? - How do the smaller angles at \(B\) combine to make \(\beta\)?

1. Since \(g\parallel AC\), the angle between \(BD\) and \(g\) at \(B\) matches \(\angle A=50^\circ\) by corresponding angles. 2. The angle between \(g\) and \(BC\) at \(B\) matches \(\angle C=75^\circ\) by alternate interior angles. 3. These adjacent angles make the exterior angle \(\beta\). Thus \(\beta=50^\circ+75^\circ=125^\circ\).

The parallel line splits \(\beta\) into angles equal to \(\angle A\) and \(\angle C\), by corresponding and alternate interior angles. So \(\beta=125^\circ\).

- The new line at \(C\) has the same direction as side \(AB\). - Compare the two angles at the base with angles at \(C\). - What sum do the three adjacent angles at \(C\) form?

1. Because \(g\parallel AB\), the angles at \(C\) between \(g\) and \(CA\), and between \(CB\) and \(g\), match \(\alpha\) and \(\beta\) by alternate interior angles. 2. Those two angles and \(\gamma\) lie beside one another on a straight line through \(C\). 3. Their sum is \(180^\circ\), so \(\alpha+\beta+\gamma=180^\circ\).

The two alternate interior angles at \(C\) match \(\alpha\) and \(\beta\); with \(\gamma\), they form a straight angle of \(180^\circ\).

- Use \(AD\parallel BC\) first and treat the straight line through \(A\), \(B\), and \(E\) as a transversal. - Then use \(AB\parallel CD\) with transversal \(BC\). - Connect the two congruence statements through the common angle at \(B\).

1. Since \(AD\parallel BC\) and line \(ABE\) is a transversal, \(\angle DAB\) and \(\angle EBC\) are corresponding angles. Therefore, \(\angle DAB\cong\angle EBC\). 2. Since \(AB\parallel CD\) and \(BC\) is a transversal, \(\angle EBC\) and \(\angle BCD\) are alternate interior angles. Therefore, \(\angle EBC\cong\angle BCD\). 3. By transitivity of angle congruence, \(\angle DAB\cong\angle BCD\).

1. Corresponding Angles Theorem 2. Alternate Interior Angles Theorem 3. \(\angle DAB\cong\angle BCD\)

- Treat diagonal \(AC\) as a transversal for both side pairs. - Identify the opposite sides proved parallel by each given angle equality. - What defines a parallelogram?

1. Angles \(BAC\) and \(ACD\) are alternate interior angles along transversal \(AC\). Their equality gives \(AB\parallel CD\). 2. Angles \(CAD\) and \(ACB\) are also alternate interior angles. Their equality gives \(AD\parallel BC\). 3. Both pairs of opposite sides are parallel, so \(ABCD\) is a parallelogram.

The two alternate interior angle converses give \(AB\parallel CD\) and \(AD\parallel BC\), so \(ABCD\) is a parallelogram.

- What does the pair of angles along diagonal \(AC\) tell you? - Which pair of opposite sides has a second useful property? - Recall a parallelogram test that needs only one side pair.

1. The equal alternate interior angles \(BAC\) and \(ACD\) give \(AB\parallel CD\) by the converse theorem. 2. The problem also states \(AB=CD\). 3. A quadrilateral with one pair of opposite sides both parallel and congruent is a parallelogram.

Since \(AB\parallel CD\) by the alternate-interior converse and \(AB=CD\), \(ABCD\) is a parallelogram.

- Begin with the interior-angle sum of a quadrilateral. - Substitute the two opposite-angle equalities. - Which supplementary pairs share a side used as a transversal?

1. The four interior angles add to \(360^\circ\). Because opposite angles match, \(2m\angle A+2m\angle B=360^\circ\). 2. Thus \(m\angle A+m\angle B=180^\circ\); also \(m\angle B+m\angle C=180^\circ\). 3. The same-side interior converse gives \(AD\parallel BC\) from the first pair and \(AB\parallel CD\) from the second. 4. Both opposite side pairs are parallel, so the quadrilateral is a parallelogram.

Equal opposite angles force both adjacent angle sums to be \(180^\circ\). The same-side interior converse proves both pairs of opposite sides parallel.

- The proof is establishing the parallelogram test, so you may not invoke that test as a shortcut. - Use the given parallel sides to obtain an angle pair involving diagonal \(AC\). - After triangle congruence, look for a corresponding angle pair that can prove the other sides parallel.

1. Draw diagonal \(\overline{AC}\). 2. Since \(AB\parallel CD\), \(\angle BAC\cong\angle ACD\) by the alternate interior angles theorem. 3. In \(\triangle ABC\) and \(\triangle CDA\), \(AB=CD\) is given, \(AC=CA\) by the reflexive property, and the included angles are congruent. 4. Therefore, \(\triangle ABC\cong\triangle CDA\) by SAS. 5. CPCTC gives \(\angle BCA\cong\angle CAD\). 6. These are alternate interior angles for lines \(BC\) and \(AD\) cut by transversal \(AC\). By the converse of the alternate interior angles theorem, \(BC\parallel AD\). 7. Now both pairs of opposite sides are parallel, so \(ABCD\) is a parallelogram.

Diagonal \(AC\) creates triangles \(ABC\) and \(CDA\). The given parallel sides supply one included-angle pair, so SAS proves the triangles congruent. CPCTC supplies a second alternate-interior-angle pair, which proves \(BC\parallel AD\). Therefore, both pairs of opposite sides are parallel and \(ABCD\) is a parallelogram.

- Which two opposite sides are cut by side \(PQ\)? - What do the supplementary interior angles tell you about them? - Combine that result with the given equal side lengths.

1. Angles \(SPQ\) and \(PQR\) are same-side interior angles formed by transversal \(PQ\) and sides \(PS\) and \(QR\). 2. Because they add to \(180^\circ\), the converse theorem gives \(PS\parallel QR\). 3. This side pair is both congruent and parallel. By the parallelogram test, \(PQRS\) is a parallelogram.

The supplementary angles give \(PS\parallel QR\). Since \(PS=QR\) as well, \(PQRS\) is a parallelogram.

- Use alternate interior angles formed by the diagonal and each pair of parallel sides. - Identify the side between the two equal angles.

1. Since \(AB \parallel CD\), alternate interior angles satisfy \(\angle BAC = \angle DCA\). 2. Since \(BC \parallel AD\), alternate interior angles satisfy \(\angle BCA = \angle DAC\). 3. The triangles share the included side \(\overline{AC}\). 4. Therefore, \(\triangle ABC \cong \triangle CDA\) by ASA.

The two pairs of alternate interior angles are congruent, and \(AC\) is the included shared side. Therefore, \(\triangle ABC \cong \triangle CDA\) by ASA.

- Which angles are equal in triangle \(DBE\)? - Transfer them across the two parallel segments. - What does equality of the larger triangle's base angles show?

1. Because \(BD=BE\), triangle \(DBE\) has equal base angles: \(\angle BDE=\angle DEB\). 2. The parallel lines \(DE\) and \(AC\) give \(\angle BAC=\angle BDE\) and \(\angle BCA=\angle DEB\). 3. Therefore, \(\angle BAC=\angle BCA\). 4. Equal base angles in triangle \(ABC\) imply \(AB=BC\).

The smaller isosceles triangle has equal base angles; parallel-line angle relationships transfer them to \(\angle BAC\) and \(\angle BCA\), so \(AB=BC\).

- Use the order of the congruence statement to identify corresponding angles. - How are those angles positioned relative to \(KL\), \(MN\), and transversal \(KM\)?

1. The congruence statement gives the correspondence \(K \leftrightarrow M\), \(L \leftrightarrow N\), and \(M \leftrightarrow K\). 2. Therefore, corresponding angles \(\angle LKM\) and \(\angle NMK\) are congruent. 3. These are alternate interior angles formed by lines \(KL\) and \(MN\) with transversal \(KM\). 4. By the converse of the alternate interior angles theorem, \(KL \parallel MN\).

Yes. Since \(\angle LKM \cong \angle NMK\), the converse of the alternate interior angles theorem gives \(KL \parallel MN\).

- Use the midpoint to identify the side included between the relevant angles. - Find one angle pair from the intersection at \(M\). - Find another angle pair from the parallel bases.

1. Since \(M\) is the midpoint of \(\overline{CD}\), \(CM=DM\). 2. Angles \(\angle BMC\) and \(\angle FMD\) are vertical angles, so they are congruent. 3. Because \(BC\parallel AF\), angles \(\angle BCM\) and \(\angle FDM\) are alternate interior angles, so they are congruent. 4. Therefore, \(\triangle BCM\cong\triangle FDM\) by ASA. 5. By CPCTC, \(BC=DF\).

a) \(\triangle BCM\cong\triangle FDM\) by ASA. b) CPCTC gives \(BC=DF\).

- Relate \(AB\), \(CD\), and \(EF\) using the two given parallelograms. - Which parallelogram test uses one pair of opposite sides? - Add the two collinear segments to find \(AF\) for the perimeter.

1. Opposite sides of \(ABCD\) give \(AB\parallel CD\) and \(AB=CD\). 2. Opposite sides of \(DCEF\) give \(CD\parallel EF\) and \(CD=EF\). 3. Thus \(AB\parallel EF\) and \(AB=EF\), so \(ABEF\) is a parallelogram by the one-side-pair test. 4. Since \(A,D,F\) are collinear in order, \(AF=AD+DF=5+4=9\,\text{cm}\). 5. Its perimeter is \(2(5+9)=28\,\text{cm}\).

a) Opposite sides \(AB\) and \(EF\) are parallel and congruent, so \(ABEF\) is a parallelogram. b) \(28\,\text{cm}\)

- Find a pair of angles equal because \(AD\parallel BC\). - Combine that equality with the angle-bisector condition. - What does the midpoint tell you about the full length \(AD\)?

1. Since \(AD\parallel BC\), alternate interior angles give \(\angle AMB=\angle MBC\). 2. Because \(BM\) bisects \(\angle ABC\), \(\angle ABM=\angle MBC\). 3. Hence \(\angle ABM=\angle AMB\), so \(AB=AM\) in isosceles triangle \(ABM\). 4. Point \(M\) is the midpoint of \(AD\), so \(AD=2AM=2AB\). 5. Opposite sides of a parallelogram are equal, giving \(BC=AD=2AB\).

Alternate interior angles and the bisector make \(AB=AM\). Since \(AD=2AM\) and \(BC=AD\), \(BC=2AB\).

- Use one pair of congruent opposite sides. - Find alternate interior angles made by each diagonal. - Identify the congruence criterion from two angles and their included side.

1. Opposite sides of a parallelogram are congruent, so \(AB = CD\). 2. Since \(AB \parallel CD\), transversal \(\overline{AC}\) gives \(\angle BAS = \angle DCS\), and transversal \(\overline{BD}\) gives \(\angle ABS = \angle CDS\). 3. The equal side lies between the two pairs of equal angles. 4. Therefore, \(\triangle ABS \cong \triangle CDS\) by ASA.

\(\triangle ABS \cong \triangle CDS\) by ASA, using opposite sides of a parallelogram and alternate interior angles formed by parallel lines.

- Calculate an interior angle of the regular pentagon. - Use the two equal sides of triangle \(ABC\). - Look for a supplementary angle pair along side \(CD\).

1. Each interior angle of a regular pentagon is \(108^\circ\). 2. Since \(AB=BC\), triangle \(ABC\) is isosceles. Its base angles measure \((180^\circ-108^\circ)\div2=36^\circ\). 3. At \(C\), \(\angle ACD=108^\circ-36^\circ=72^\circ\). 4. Since \(\angle CDE=108^\circ\), the same-side interior angles along transversal \(CD\) sum to \(72^\circ+108^\circ=180^\circ\). 5. The same-side interior converse gives \(AC\parallel DE\).

a) \(\angle ABC=108^\circ\) and \(\angle BAC=36^\circ\). b) \(\angle ACD=72^\circ\) and \(\angle CDE=108^\circ\) are supplementary, so \(AC\parallel DE\).

- Classify the two marked angles by their positions before using their measures. - A converse theorem has hypotheses about both angle measure and angle position. - Check whether Samira verified every hypothesis of the theorem named.

1. The measures are supplementary because \(80^\circ+100^\circ=180^\circ\). 2. However, \(\angle\alpha\) and \(\angle\beta\) occupy corresponding positions in the diagram; they are not same-side interior angles. 3. The Same-Side Interior Angles Converse applies only when the supplementary pair is a same-side interior pair. 4. Therefore, Samira's argument is invalid, and the given supplementary corresponding angles do not establish \(g\parallel h\).

No. The angles are corresponding, not same-side interior, so the Same-Side Interior Angles Converse does not apply. Parallelism does not follow from the stated fact.

- The target angles are at different intersections; look for a third angle that can connect them. - At \(P\), identify an angle vertical to \(\angle APQ\). - Compare that third angle with \(\angle PQD\) using the parallel lines.

1. At the upper intersection, \(\angle APQ\) and \(\angle EPB\) are vertical angles, so \(\angle APQ\cong\angle EPB\). 2. Because \(g\parallel h\), \(\angle EPB\) and \(\angle PQD\) are corresponding angles, so \(\angle EPB\cong\angle PQD\). 3. By transitivity of angle congruence, \(\angle APQ\cong\angle PQD\).

Use \(\angle EPB\) as a bridge: \(\angle APQ\cong\angle EPB\) by vertical angles and \(\angle EPB\cong\angle PQD\) by corresponding angles, so \(\angle APQ\cong\angle PQD\) by transitivity.

- Introduce an angle at \(Q\) that is alternate interior to \(\angle APQ\). - At \(Q\), look for a linear pair involving that auxiliary angle and \(\angle CQP\). - Replace one measure in the linear-pair equation with an equal measure from the parallel-line relationship.

1. \(\angle APQ\) and \(\angle PQD\) are alternate interior angles, so \(m\angle APQ=m\angle PQD\). 2. \(\angle PQD\) and \(\angle CQP\) form a linear pair, so \(m\angle PQD+m\angle CQP=180^\circ\). 3. Substitute \(m\angle APQ\) for \(m\angle PQD\). Therefore, \(m\angle APQ+m\angle CQP=180^\circ\).

\(\angle APQ\cong\angle PQD\) by alternate interior angles, while \(\angle PQD\) and \(\angle CQP\) form a linear pair. Substitution gives \(m\angle APQ+m\angle CQP=180^\circ\).

- The given supplementary equation does not itself use a standard same-side interior pair. - Step 1 lets you replace one angle measure in the given equation with an equal measure. - The missing line must create exactly the hypothesis needed for the converse theorem used in Step 4.

1. Vertical angles are congruent, so \(m\angle EPB=m\angle APQ\). 2. Substitute \(m\angle APQ\) for \(m\angle EPB\) in the given equation. This gives \(m\angle APQ+m\angle CQP=180^\circ\). 3. The angles \(\angle APQ\) and \(\angle CQP\) are same-side interior angles. 4. Since that same-side interior pair is supplementary, the converse theorem gives \(g\parallel h\).

Step 2: \(m\angle APQ+m\angle CQP=180^\circ\), by substitution using \(m\angle EPB=m\angle APQ\).

- Replace the conclusion you want with an auxiliary line through \(Q\) whose parallelism to \(g\) is guaranteed. - Compare the angle made by that auxiliary line with \(t\) to the angle made by \(h\) with \(t\). - Use the stated uniqueness facts to show the auxiliary line cannot be different from \(h\).

1. Through \(Q\), draw line \(k\) parallel to \(g\). 2. Because \(g\parallel k\), the Corresponding Angles Theorem says that the corresponding angle formed by \(k\) and \(t\) at \(Q\) is congruent to the given angle formed by \(g\) and \(t\) at \(P\). 3. By the original given, the corresponding angle formed by \(h\) and \(t\) at \(Q\) is congruent to that same angle at \(P\). Therefore, the rays of \(h\) and \(k\) in the corresponding position make equal angles with the same ray of \(t\), on the same side. 4. By uniqueness of a ray with a specified angle from a fixed ray, those rays coincide. Hence lines \(h\) and \(k\) are the same line. 5. Since \(k\parallel g\) and \(h=k\), it follows that \(g\parallel h\).

Draw the unique line \(k\) through \(Q\) parallel to \(g\). The Corresponding Angles Theorem makes the relevant angle for \(k\) equal to the given corresponding angle at \(P\); the hypothesis makes the relevant angle for \(h\) equal to that same angle. Ray uniqueness forces \(h=k\), so \(g\parallel h\).

- First identify which statements are raw givens or reflexive facts and which require triangle congruence. - A CPCTC statement cannot be justified until the triangle-congruence statement has been established. - Each parallel-line conclusion needs the matching angle congruence before it. - The final quadrilateral classification needs both parallel-side conclusions.

1. Start with E, the given side congruences. 2. Use D, the reflexive fact \(AC=CA\). 3. From E and D, the three corresponding side pairs are congruent, so C follows by SSS. 4. From C, CPCTC gives both A and F. These two statements may appear in either order. 5. From A, the converse of the alternate interior angles theorem gives B. 6. From F, the converse of the alternate interior angles theorem gives G. 7. Once B and G are established, H follows from the definition of a parallelogram. 8. One valid order is E, D, C, A, B, F, G, H.

One valid order is E, D, C, A, B, F, G, H. D: reflexive property. C: SSS. A: CPCTC. B: converse of the alternate interior angles theorem. F: CPCTC. G: converse of the alternate interior angles theorem. H: both pairs of opposite sides are parallel, so \(ABCD\) is a parallelogram. A and F may be interchanged, and the two resulting parallel-line deductions may be interleaved as long as each comes after its required CPCTC statement.

- Can the first pair of measurements by itself establish parallelism? - If the bars are parallel, what must be true at every other transversal? - Which two pieces of evidence disagree?

1. The angles on the first support add to \(112^\circ+68^\circ=180^\circ\), so the bars must be parallel if those measurements and angle positions are exact. 2. For parallel bars, corresponding angles on the second support must be equal. But \(54^\circ\ne57^\circ\). 3. Therefore the measurements cannot all be exact under the stated conditions. The crew should recheck the angle measurements or which angles were identified as corresponding.

No. The first support proves parallelism, but the second gives unequal corresponding angles. At least one measurement or angle identification must be wrong.

- Which two smaller lengths would help you compare triangle \(BKP\) with triangle \(ABC\)? - What feature do the two triangles share at \(B\)? - What kind of angle relationship can establish that two lines are parallel?

1. Subtract equal parts from the equal sides: \(BK=AB-AK=BC-CP=BP\). 2. Thus triangle \(BKP\) is isosceles and shares vertex angle \(B\) with isosceles triangle \(ABC\). 3. Both triangles have the same two base-angle measures, so \(\angle BKP=\angle BAC\). 4. These corresponding angles along transversal \(AB\) are equal. The corresponding-angle converse gives \(KP\parallel AC\).

Yes. \(BK=BP\), so triangles \(BKP\) and \(ABC\) have matching base angles, which establish \(KP\parallel AC\) by a converse angle theorem.

- Try extending one base and constructing a segment equal in length to the other base. - What quadrilateral is formed when one pair of opposite sides is both parallel and congruent? - Use the auxiliary parallelogram and the congruent diagonals to create an angle equality for an SAS proof.

1. On ray \(AB\) beyond \(B\), construct point \(E\) so that \(BE = CD\). 2. Since \(BE \parallel CD\) and \(BE = CD\), quadrilateral \(BECD\) is a parallelogram. Therefore, \(EC \parallel BD\) and \(EC = BD\). 3. The given \(AC = BD\), together with \(EC = BD\), gives \(AC = EC\). Thus, \(\triangle ACE\) is isosceles and \(\angle CAE = \angle AEC\). 4. Because \(A\), \(B\), and \(E\) are collinear, \(\angle CAE = \angle CAB\). Because \(EA\) is collinear with \(BA\) and \(EC \parallel BD\), \(\angle AEC = \angle DBA\). Hence \(\angle CAB = \angle DBA\). 5. In \(\triangle CAB\) and \(\triangle DBA\), \(CA = DB\), \(AB = BA\), and the included angles \(\angle CAB\) and \(\angle DBA\) are congruent. Therefore, \(\triangle CAB \cong \triangle DBA\) by SAS. 6. Corresponding sides give \(CB = DA\). Therefore, \(ABCD\) is an isosceles trapezoid.

Extend ray \(AB\) beyond \(B\) and choose \(E\) so that \(BE=CD\). Then \(BECD\) is a parallelogram, leading to \(AC=EC\) and \(\angle CAB=\angle DBA\). Thus \(\triangle CAB\cong\triangle DBA\) by SAS, so \(AD=BC\).

- How are opposite angles of a parallelogram related? - Compare half of each angle against corresponding parallel sides. - Where does the condition “not a rhombus” enter your reasoning?

1. Opposite interior angles \(A\) and \(C\) are equal, so their halves have equal measures. 2. The bisectors make equal corresponding angles with the parallel sides \(AB\) and \(CD\); hence \(p\) and \(q\) have the same direction. 3. If \(p\) and \(q\) were the same line, it would be diagonal \(AC\). Then \(\angle BAC=\angle CAD\). Since \(AD\parallel BC\), \(\angle CAD=\angle ACB\). 4. Thus triangle \(ABC\) would have equal angles at \(A\) and \(C\), giving \(AB=BC\). A parallelogram with adjacent sides equal is a rhombus, contrary to the given condition. 5. Therefore \(p\) and \(q\) are distinct parallel lines.

Equal opposite angles give equally directed bisectors, and the non-rhombus condition prevents them from coinciding. Thus \(p\parallel q\).

- Which groups of four vertices might form a parallelogram? - What lengths can you compare using opposite sides of a parallelogram? - How could the equal lengths help compare triangle areas without using similarity?

1. Since \(AB\parallel CY\) and \(AY\parallel BC\), quadrilateral \(ABCY\) is a parallelogram. Thus \(AY=BC\). 2. Since \(AZ\parallel BC\) and \(BZ\parallel AC\), quadrilateral \(AZBC\) is a parallelogram. Thus \(AZ=BC\) and \(ZB=AC\). 3. Since \(AB\parallel XC\) and \(BX\parallel AC\), quadrilateral \(ABXC\) is a parallelogram. Thus \(XB=AC\) and \(XC=AB\). It follows that \(AY=AZ\), \(ZB=BX\), and \(XC=CY\). Each point lies between the endpoints of its side, so \(A\), \(B\), and \(C\) are the midpoints. 4. By the three parallelograms, each corner triangle \(YAC\), \(ZAB\), and \(XBC\) has the three side lengths of triangle \(ABC\). The four triangles are congruent by SSS. 5. They fill \(XYZ\) without overlap, so \([XYZ]:[ABC]=4:1\).

a) \(ABCY\), \(AZBC\), and \(ABXC\) are parallelograms. Their equal opposite sides give \(AY=AZ\), \(ZB=BX\), and \(XC=CY\), so \(A\), \(B\), and \(C\) are midpoints. b) Triangles \(YAC\), \(ZAB\), \(XBC\), and \(ABC\) are congruent by SSS. They fill \(XYZ\), so the area ratio \([XYZ]:[ABC]\) is \(4:1\).