- The compass opening should stay fixed after it is matched to the source segment. - Each new endpoint becomes the center for the next transfer. - Compare how many copied intervals lie on each side of \(P_2\).

1. Set the compass opening to \(AB\) and do not change it. 2. Center the compass at \(P_0\) and mark \(P_1\) on the ray. Then center it successively at \(P_1\), \(P_2\), and \(P_3\) to mark \(P_2\), \(P_3\), and \(P_4\). 3. The four consecutive copied segments each have length \(2.35\,\text{cm}\), so \(P_0P_4=4\cdot2.35=9.40\,\text{cm}\). 4. Each of \(P_0P_2\) and \(P_2P_4\) consists of two copied segments, so both equal \(4.70\,\text{cm}\). Therefore, \(P_2\) is the midpoint.

Keep the compass set to \(AB\). Center successively at \(P_0,P_1,P_2,P_3\) to mark the next point on the ray each time. Then \(P_0P_4=9.40\,\text{cm}\), and \(P_2\) is the midpoint because \(P_0P_2=P_2P_4=4.70\,\text{cm}\).

- First create a copy of the longer source segment on the target ray. - For the subtraction step, think about which endpoint should be the center of the second arc. - The second copied segment must fit inside the first copied segment for the requested point placement.

1. Set the compass opening to \(AB\). With center \(P\), mark its intersection \(Q\) with ray \(\overrightarrow{PX}\), so \(PQ=AB\). 2. Reset the compass opening to \(CD\). With center \(Q\), draw an arc that meets \(\overline{PQ}\) at \(R\). 3. Then \(RQ=CD\), so \(PR=PQ-RQ=AB-CD\). 4. Numerically, \(PR=9.2-3.7=5.5\,\text{cm}\). 5. The required intersection lies on \(\overline{PQ}\) when \(CD\le AB\).

First copy \(AB\) from \(P\) to locate \(Q\) on \(\overrightarrow{PX}\). Then set the compass to \(CD\), center it at \(Q\), and use the intersection of that arc with \(\overline{PQ}\) as \(R\). This gives \(PR=AB-CD=5.5\,\text{cm}\). The construction has \(R\in\overline{PQ}\) when \(CD\le AB\).

- A standard angle copy transfers more than just one compass opening. - First match the source and target arc radii; then consider the chord cut off by the source angle. - The two copies are placed on opposite sides of the same initial ray.

1. Draw an arc centered at the source vertex so that it meets both sides of the source angle. Preserve that arc radius. 2. With the same radius, draw target arcs centered at \(P\) on the two sides of \(\overrightarrow{PX}\). 3. Measure with the compass the chord between the two source-arc intersections. Transfer that chord from the point where each target arc meets \(\overrightarrow{PX}\), choosing opposite sides of the ray, and draw \(\overrightarrow{PU}\) and \(\overrightarrow{PV}\) through the transferred points. 4. Each copied angle is \(52^\circ\), so the smaller angle between the terminal rays is \(52^\circ+52^\circ=104^\circ\). 5. Ray \(\overrightarrow{PX}\) divides \(\angle UPV\) into two congruent angles, so it is the angle bisector.

Transfer the source arc radius and then the source chord between the arc intersections to construct the two copies on opposite sides of \(\overrightarrow{PX}\). The smaller angle is \(104^\circ\), and \(\overrightarrow{PX}\) is its angle bisector.

- Every copied segment must start from the specified point \(P\). - The compass must match the appropriate source segment before each transfer. - Decide whether \(P\) lies between the endpoints before choosing addition or subtraction.

1. Set the compass to the first source segment. Center it at \(P\) and mark its intersection with one ray as \(U\). 2. Reset the compass to the second source segment. Center it at \(P\) and mark its intersection with the opposite ray as \(V\). 3. Since \(P\) lies between \(U\) and \(V\), \(UV=PU+PV=4.8+7.1=11.9\,\text{cm}\). 4. For same-ray placement, center both transfers at \(P\) on one ray. The endpoints are then separated by \(7.1-4.8=2.3\,\text{cm}\).

Copy \(4.8\,\text{cm}\) from \(P\) onto one ray, reset the compass to \(7.1\,\text{cm}\), and copy that length from \(P\) onto the opposite ray. Then \(UV=11.9\,\text{cm}\). On the same ray, both copies start at \(P\), and their endpoints are \(2.3\,\text{cm}\) apart.

- Translate each point's membership on a circle into a distance equality. - Track whether the compass opening changes between the two circles. - Compare the three side lengths of the resulting triangle.

1. Point \(Q\) lies on the circle centered at \(P\) with radius \(AB\), so \(PQ=AB\). 2. Point \(R\) lies on the circle centered at \(P\) with the same radius, so \(PR=AB\). 3. Point \(R\) also lies on the circle centered at \(Q\) with radius \(AB\), so \(QR=AB\). 4. Thus \(PQ=PR=QR\), so \(\triangle PQR\) is equilateral.

All three sides are copies of \(\overline{AB}\): \(PQ=PR=QR=AB\). Therefore, \(\triangle PQR\) is equilateral.

- Compare the radii and chord lengths in the source and target constructions. - Think about the congruence of the triangles formed by the arc centers and intersection points. - Use the stated direction of rotation to select the required side.

1. The target arc has the same radius as the source arc. 2. The transferred chord has the same length as the chord cut off by the source angle. 3. For either intersection point, the two radii and the transferred chord form a triangle congruent to the source construction triangle by SSS. 4. Therefore, both \(\angle QPR\) and \(\angle QPS\) have the same measure as the source angle, but they lie on opposite sides of the target ray. 5. Amina must choose the intersection reached by rotating counterclockwise from \(\overrightarrow{PQ}\).

Both intersections produce congruent construction triangles, so both angles are valid copies. To satisfy the orientation condition, choose the point on the counterclockwise side of \(\overrightarrow{PQ}\).

- Recall the two compass quantities used in a standard angle-copy construction. - The phrase “same side” determines which of the possible copied rays to use. - After the construction, use the two copied base angles to analyze the triangle.

1. For each copy, use the standard angle-copy construction: transfer the source arc radius to the new vertex, then transfer the chord between the two source-arc intersections to locate the terminal ray. 2. Choose the terminal rays at \(P\) and \(Q\) on the same side of \(\overline{PQ}\); their intersection is \(R\). 3. The copied base angles satisfy \(\angle RPQ=\angle PQR=55^\circ\). 4. Equal base angles in a triangle have opposite congruent sides, so \(PR=QR\). 5. The triangle angle sum gives \(m\angle PRQ=180^\circ-55^\circ-55^\circ=70^\circ\).

Each angle copy transfers the source arc radius and then the source chord. With the two copies opening to the same side of \(\overline{PQ}\), their rays meet at \(R\). The equal copied base angles give \(PR=QR\), and \(m\angle PRQ=70^\circ\).

- Think of each copied length as a fixed distance from one endpoint of \(\overline{AC}\). - Use loci that contain all points at each required distance. - Check what side equalities two common circle intersections automatically create.

1. Draw a circle centered at \(A\) with radius equal to the copied \(5\,\text{cm}\) segment. 2. Draw a circle centered at \(C\) with radius equal to the copied \(4\,\text{cm}\) segment. 3. The circles intersect at two points because \(|5-4|<6<5+4\). Label the two intersections \(B\) and \(D\). 4. By the circle radii, \(AB=AD=5\,\text{cm}\) and \(CB=CD=4\,\text{cm}\). 5. Thus quadrilateral \(ABCD\) has two distinct pairs of congruent consecutive sides and is a kite.

Intersect a circle centered at \(A\) with radius \(5\,\text{cm}\) and a circle centered at \(C\) with radius \(4\,\text{cm}\). Use the two intersections as \(B\) and \(D\). The radius equalities give the two pairs of congruent consecutive sides required for a kite.

- Distinguish copying a distance from \(P\) from appending a new distance to an existing copied segment. - The first endpoint should become relevant to the placement of the second copied segment. - Once each construction is complete, compare the final distance from \(P\).

1. In the first construction, set the compass to \(a\), center it at \(P\), and mark the first endpoint on the ray. Then reset the compass to \(b\), center it at that new endpoint, and mark the final endpoint farther along the ray. 2. In the second construction, set the compass to \(b\), center it at \(P\), and mark the first endpoint. Reset to \(a\), center at that endpoint, and mark the final endpoint. 3. Centering the second transfer again at \(P\) would merely mark another distance from \(P\); it would not append the second length to the first segment. 4. The first final endpoint is at distance \(a+b\) from \(P\), while the second is at distance \(b+a\). 5. Since \(a+b=b+a\), the two points lie at the same distance from \(P\) on the same ray, so they coincide.

For order \(a\) then \(b\), center the \(a\)-transfer at \(P\) and the \(b\)-transfer at the first endpoint. Reverse the openings for order \(b\) then \(a\). The second transfer must start at the first endpoint to append the lengths. The final distances are \(a+b\) and \(b+a\), so the final endpoints coincide.

- Reproduce the direction of \(\overline{AB}\) symmetrically across \(m\). - Preserve the distance from \(A\). - Use the properties of an isosceles triangle to verify reflection symmetry.

1. Copy the angle between ray \(\overrightarrow{AB}\) and line \(m\) to the opposite side of \(m\), using the ray of \(m\) from \(A\) as the common reference side. 2. On the copied ray, transfer length \(AB\) to locate point \(B''\), so \(AB''=AB\). 3. The construction gives congruent angles between \(m\) and \(\overline{AB}\), and between \(m\) and \(\overline{AB''}\), on opposite sides of \(m\). 4. Thus line \(m\) bisects \(\angle BAB''\), and \(\triangle ABB''\) is isosceles with \(AB=AB''\). 5. In an isosceles triangle, the vertex-angle bisector is the perpendicular bisector of the base. Therefore, \(m\) is the perpendicular bisector of \(\overline{BB''}\), so \(B''\) is the reflection of \(B\).

Copy the angle made by \(\overline{AB}\) and \(m\) to the opposite side of \(m\), then copy length \(AB\) on the new ray. The resulting point \(B''\) is the reflection because \(m\) is the perpendicular bisector of \(\overline{BB''}\).

- Start by comparing the longest segment with the other two placed end to end. - A common ray lets you compare the copied lengths without numerical measurement. - For the construction, think of each remaining side length as a fixed-distance condition from one endpoint of the base.

1. Copy the \(4.5\,\text{cm}\) and \(6\,\text{cm}\) segments consecutively on one ray. Their combined length is \(10.5\,\text{cm}\). 2. Copy the \(9.8\,\text{cm}\) segment from the same starting point on the same ray. Its endpoint lies before the \(10.5\,\text{cm}\) endpoint, so \(4.5+6>9.8\). 3. Because \(9.8\,\text{cm}\) is the longest length, the other two triangle inequalities hold automatically. The three lengths can form a triangle. 4. Copy the \(9.8\,\text{cm}\) segment as base \(\overline{AB}\). 5. Draw a circle centered at \(A\) with radius equal to the \(4.5\,\text{cm}\) source segment and a circle centered at \(B\) with radius equal to the \(6\,\text{cm}\) source segment. 6. Either circle intersection can be used as the third vertex \(C\); then connect \(C\) to \(A\) and \(B\).

Yes. The copied lengths show \(4.5\,\text{cm}+6\,\text{cm}=10.5\,\text{cm}>9.8\,\text{cm}\). Copy \(9.8\,\text{cm}\) as the base, then intersect circles centered at its endpoints with radii copied from the \(4.5\,\text{cm}\) and \(6\,\text{cm}\) source segments.

- Begin by creating two rays perpendicular to \(\overline{AB}\) on the same side. - Transfer the given side length onto both perpendicular rays. - Before calling the quadrilateral a square, identify a theorem that first proves it is a parallelogram.

1. Copy the right angle at \(A\) with one side along \(\overrightarrow{AB}\), and copy it at \(B\) with one side along \(\overrightarrow{BA}\), both opening to the chosen side of \(\overline{AB}\). 2. Copy length \(AB\) onto the perpendicular ray at \(A\) to locate \(D\), and onto the perpendicular ray at \(B\) to locate \(C\). 3. Since \(AD\perp AB\) and \(BC\perp AB\), lines \(AD\) and \(BC\) are parallel. 4. Also, \(AD=BC=AB\). A quadrilateral with one pair of opposite sides both parallel and congruent is a parallelogram, so \(ABCD\) is a parallelogram. 5. Therefore \(CD=AB\). The parallelogram has four congruent sides and a right angle, so it is a square.

Copy right angles at \(A\) and \(B\) on the same side of \(\overline{AB}\), copy length \(AB\) along both new rays to locate \(D\) and \(C\), and join \(C\) to \(D\). The result is a parallelogram with four congruent sides and a right angle, so it is a square.

- Identify the three side lengths used in the usual SSS justification. - Two of those lengths are already correct because the arc radius was preserved. - Determine which remaining length the changed compass opening affects.

1. Equal source and target arc radii correctly create two equal sides in the source and target construction triangles. 2. The third required equality is the chord between the two arc-intersection points. 3. Using a longer compass opening makes the target chord different from the source chord. 4. The two construction triangles are no longer guaranteed congruent by SSS, so the target angle is not guaranteed congruent to the source angle.

The chord-length condition fails: the target chord is not equal to the source chord. Without that third side equality, SSS does not guarantee equal angles.

- A compass circle can represent all points at a fixed distance from one endpoint of the diagonal. - Use the known diagonal as the common baseline for locating both remaining vertices. - Think about how two circle intersections lie relative to the line through their centers.

1. Draw \(\overline{AC}\) with length \(7\,\text{cm}\). 2. Draw a circle centered at \(A\) with radius \(6\,\text{cm}\) and a circle centered at \(C\) with radius \(5\,\text{cm}\). Choose either intersection as \(B\). 3. Draw a circle centered at \(A\) with radius \(3\,\text{cm}\) and a circle centered at \(C\) with radius \(5\,\text{cm}\). 4. Since \(|5-3|<7<5+3\), the second pair of circles has two intersections, one on each side of \(\overline{AC}\). These are the two possible locations for \(D\). 5. Once \(B\) is fixed, choose \(D\) on the opposite side of \(\overline{AC}\) from \(B\) to obtain the convex quadrilateral. The other location produces the concave configuration.

Construct \(B\) as an intersection of circles centered at \(A\) and \(C\) with radii \(6\,\text{cm}\) and \(5\,\text{cm}\). Construct \(D\) as an intersection of circles centered at \(A\) and \(C\) with radii \(3\,\text{cm}\) and \(5\,\text{cm}\). After \(A\), \(B\), and \(C\) are fixed, there are two possible locations for \(D\); the location on the opposite side of \(AC\) from \(B\) gives the convex quadrilateral.

- Build the figure one vertex at a time rather than trying to place all four vertices at once. - After \(C\) is placed, compare the two possible rays for \(CD\) with the side of line \(BC\) that contains \(A\). - Distinguish a local choice that affects convexity from reflecting the entire finished figure.

1. Draw \(\overline{AB}\) with the given length. 2. At \(B\), copy the given angle \(\angle B\). On the chosen ray, mark \(C\) so that \(BC\) has the given length. 3. At \(C\), copy the given angle \(\angle C\). Choose the ray that places \(D\) on the same side of line \(BC\) as \(A\), as required for the convex orientation. 4. On that ray, mark \(D\) so that \(CD\) has the given length, then connect \(D\) to \(A\). 5. Once \(A\), \(B\), and the side of \(\overline{AB}\) containing \(C\) are fixed, each copied angle fixes a ray and each copied length fixes one point on that ray. At \(C\), convexity selects the ray that keeps \(A\) and \(D\) on the same side of line \(BC\). Reflecting the entire construction across line \(AB\) gives a mirror image, but that image is congruent to the first construction. Thus the quadrilateral is unique up to congruence.

a) Copy \(AB\), then copy \(\angle B\), copy \(BC\) along that ray, copy \(\angle C\) on the side of line \(BC\) that keeps \(A\) and \(D\) on the same side, and copy \(CD\) along the new ray before joining \(D\) to \(A\). b) Each copied angle fixes a ray after its side is chosen, and each copied length fixes one point on that ray. Convexity fixes the local choice at \(C\); reflecting the whole construction gives only a congruent mirror image. Therefore the result is unique up to congruence.

- Choose one copied length to serve as a fixed base. - Think about the locus of points at each specified distance from the base endpoints. - Compare the side lengths in the two possible placements.

1. Copy the \(8\,\text{cm}\) segment to make base \(\overline{PQ}\). 2. Draw a circle centered at \(P\) with radius equal to the copied \(5\,\text{cm}\) segment. 3. Draw a circle centered at \(Q\) with radius equal to the copied \(6\,\text{cm}\) segment. 4. The center distance and radii satisfy \(|6-5|<8<6+5\), so the circles intersect at two points on opposite sides of line \(PQ\). 5. Either intersection can be chosen as vertex \(R\). 6. In either position, \(PR=5\,\text{cm}\), \(QR=6\,\text{cm}\), and \(PQ=8\,\text{cm}\). The two triangles are reflections across line \(PQ\) and are congruent by SSS.

Copy the \(8\,\text{cm}\) length as \(\overline{PQ}\), then intersect a circle centered at \(P\) with radius \(5\,\text{cm}\) and a circle centered at \(Q\) with radius \(6\,\text{cm}\). Since \(|6-5|<8<6+5\), there are two possible third-vertex positions. They give reflected, congruent triangles.

- First use the copied angle to fix the directions of the two adjacent sides. - After \(B\) and \(D\) are fixed, what loci enforce the two required opposite-side lengths? - If the two loci meet twice, compare the two candidates relative to line \(BD\) and the desired uncrossed quadrilateral.

1. Copy the \(65^\circ\) angle at \(A\) to create rays \(\overrightarrow{AB}\) and \(\overrightarrow{AD}\). 2. Copy \(6\,\text{cm}\) onto \(\overrightarrow{AB}\) to locate \(B\), and copy \(4\,\text{cm}\) onto \(\overrightarrow{AD}\) to locate \(D\). 3. Draw a circle centered at \(B\) with radius \(4\,\text{cm}\) and a circle centered at \(D\) with radius \(6\,\text{cm}\). These circles have two intersections. 4. Choose as \(C\) the intersection on the side of line \(BD\) opposite \(A\). The other intersection gives a crossed configuration rather than the intended parallelogram. 5. By construction, \(AB=CD=6\,\text{cm}\) and \(AD=BC=4\,\text{cm}\). 6. A quadrilateral with both pairs of opposite sides congruent is a parallelogram, so \(ABCD\) is the required parallelogram.

Copy the angle and the two adjacent side lengths at \(A\). Then intersect the circle centered at \(B\) with radius \(4\,\text{cm}\) and the circle centered at \(D\) with radius \(6\,\text{cm}\). Choose the intersection opposite \(A\) across line \(BD\) as \(C\). This gives \(AB=CD\) and \(AD=BC\), so \(ABCD\) is a parallelogram.

- Form each new side by placing the appropriate pair of source segments end to end. - Once the three new lengths are known, choose one as a base. - Check whether the two remaining fixed-distance loci can intersect.

1. Copy the \(3\,\text{cm}\) and \(4\,\text{cm}\) segments consecutively on an auxiliary ray to create a \(7\,\text{cm}\) segment. 2. In the same way, create segments of lengths \(3+5=8\,\text{cm}\) and \(4+5=9\,\text{cm}\). 3. Copy the \(9\,\text{cm}\) segment as a base. 4. From one endpoint, draw a circle with radius equal to the copied \(7\,\text{cm}\) segment; from the other endpoint, draw a circle with radius equal to the copied \(8\,\text{cm}\) segment. 5. Since \(7+8>9\), the two circles intersect. Either intersection completes an SSS construction of the triangle.

The required side lengths are \(7\,\text{cm}\), \(8\,\text{cm}\), and \(9\,\text{cm}\). Form those lengths by copying source segments end to end, then use an SSS circle-intersection construction. The triangle can be constructed because \(7+8>9\).

- Focus first on what must be true of the diagonals at their intersection in every parallelogram. - The full diagonal lengths must be distributed symmetrically about their intersection point. - After the four vertices are placed, identify the quadrilateral theorem that certifies a parallelogram.

1. Bisect the two source segments to obtain lengths \(5\,\text{cm}\) and \(3\,\text{cm}\). 2. At a point \(O\), construct two intersecting lines making an angle of \(70^\circ\). 3. On opposite rays of the first line, locate \(A\) and \(C\) so that \(OA=OC=5\,\text{cm}\). 4. On opposite rays of the second line, locate \(B\) and \(D\) so that \(OB=OD=3\,\text{cm}\). 5. Connect \(A\), \(B\), \(C\), and \(D\) in order. Point \(O\) is the midpoint of both diagonals \(\overline{AC}\) and \(\overline{BD}\). 6. A quadrilateral whose diagonals bisect each other is a parallelogram, so \(ABCD\) is a parallelogram.

Bisect each source diagonal length, place equal halves on opposite rays of two lines meeting at \(70^\circ\), and join the four endpoints in order. The diagonals bisect each other at \(O\), so the quadrilateral is a parallelogram.

- Translate the median condition into a condition involving the midpoint of the base. - The angle condition and median-length condition describe two different loci for \(C\). - To count the forward intersections, compare the distance from the ray's endpoint to the circle center with the radius.

1. Copy \(AB=10\,\text{cm}\) and construct its midpoint \(M\). 2. At \(A\), construct the \(40^\circ\) ray on the chosen side of \(AB\). Point \(C\) must lie on this ray. 3. The condition \(CM=7\,\text{cm}\) also places \(C\) on the circle centered at \(M\) with radius \(7\,\text{cm}\). 4. Since \(AM=5\,\text{cm}<7\,\text{cm}\), point \(A\) lies inside that circle. A ray beginning at an interior point exits the circle at exactly one point. 5. That unique forward intersection is \(C\). Because \(M\) is the midpoint of \(AB\), \(\overline{CM}\) is the required median.

Construct the midpoint \(M\) of \(\overline{AB}\), the \(40^\circ\) ray from \(A\), and the circle centered at \(M\) with radius \(7\,\text{cm}\). Their unique forward intersection is \(C\). Since \(AM=5\,\text{cm}<7\,\text{cm}\), the ray starts inside the circle and exits it exactly once.

- Start from a defining property of the diagonals of a parallelogram. - Think about what lengths from the diagonal intersection to the vertices would be needed. - A point reflection can create the opposite half of a segment once its midpoint is fixed.

1. Bisect the source diagonal lengths to obtain \(5\,\text{cm}\) and \(4\,\text{cm}\). 2. Copy \(AB=6\,\text{cm}\). 3. Construct point \(O\) as an intersection of the circle centered at \(A\) with radius \(5\,\text{cm}\) and the circle centered at \(B\) with radius \(4\,\text{cm}\). 4. Reflect \(A\) across \(O\) to point \(C\), and reflect \(B\) across \(O\) to point \(D\). 5. Then \(O\) is the midpoint of both \(\overline{AC}\) and \(\overline{BD}\), with \(AC=10\,\text{cm}\) and \(BD=8\,\text{cm}\). 6. A quadrilateral whose diagonals bisect each other is a parallelogram, so \(ABCD\) has all required measurements.

Create \(\triangle AOB\) with \(AB=6\,\text{cm}\), \(AO=5\,\text{cm}\), and \(BO=4\,\text{cm}\), then reflect \(A\) and \(B\) across \(O\). The resulting diagonals bisect each other and have lengths \(10\,\text{cm}\) and \(8\,\text{cm}\), so the quadrilateral is a parallelogram.

- Compare the two base lengths and think about how their difference can be represented on the longer base. - Look for a way to enforce two distances at once when locating a new vertex. - A useful auxiliary quadrilateral may turn the required parallelism into a known quadrilateral property.

1. Copy the \(10\,\text{cm}\) segment to form base \(\overline{AB}\). 2. On \(\overline{AB}\), locate \(E\) so that \(EB=6\,\text{cm}\). Then \(AE=4\,\text{cm}\). 3. Locate \(D\) as an intersection of a circle centered at \(A\) with radius \(5\,\text{cm}\) and a circle centered at \(E\) with radius \(4\,\text{cm}\). 4. Locate \(C\), on the same side of \(\overline{AB}\) as \(D\), as an intersection of a circle centered at \(D\) with radius \(6\,\text{cm}\) and a circle centered at \(B\) with radius \(4\,\text{cm}\). 5. In quadrilateral \(DCEB\), \(DC=EB=6\,\text{cm}\) and \(CB=DE=4\,\text{cm}\). Both pairs of opposite sides are congruent, so \(DCEB\) is a parallelogram. 6. Therefore, \(DC\parallel EB\). Since \(\overline{EB}\) lies on \(\overline{AB}\), \(CD\parallel AB\), and \(ABCD\) has the required side lengths.

Copy \(AB=10\,\text{cm}\), mark \(EB=6\,\text{cm}\), locate \(D\) from the two required distances to \(A\) and \(E\), then locate \(C\) from the two required distances to \(D\) and \(B\). The auxiliary quadrilateral \(DCEB\) has both pairs of opposite sides congruent, so it is a parallelogram and \(CD\parallel AB\).