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Rule

Based on the diagram above, the following relation holds true.

Trapezoid $ABCD$ is isosceles if and only if $AC≅BD.$

To prove a biconditional statement, the conditional statement and its converse must be proven. Start by assuming that a trapezoid is isosceles.

Let $ABCD$ be an isosceles trapezoid with $AB≅CD.$ By the Isosceles Trapezoid Base Angles Theorem, the base angles are congruent, that is, $∠A≅∠D.$

Next, draw the diagonals and separate the triangles $ABD$ and $DCA.$ The Reflexive Property of Congruence gives that $AD≅AD.$
Notice that $△ABD≅△DCA$ by the Side-Angle-Side (SAS) Congruence Theorem. As a result of that relationship, $AC≅BD.$

For the converse, consider a trapezoid with congruent diagonals.

Next, draw a line parallel to $BD$ passing through $C$ and let $P$ be intersection point between this line and $AD.$

Since $AD∥BC$ and $BD∥CP,$ $BCPD$ is a parallelogram. Therefore, $CP≅BD$ and then, by the Transitive Property of Congruence, $CP≅AC.$ This makes $△ACP$ an isosceles triangle.

The Isosceles Triangle Theorem leads to the conclusion that $∠CAD≅∠CPA.$ Additionally, the Corresponding Angles Theorem indicates that $∠CPA≅∠BDA.$ Next, separate triangles $ABD$ and $DCA.$
By the Side-Angle-Side (SAS) Congruence Theorem, $△ABD≅△DCA.$ This implies that $AB≅CD$ which makes $ABCD$ an isosceles trapezoid.

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