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

If a Trapezoid Is Isosceles, Then Its Diagonals Are Congruent

Let $ABCD$ be an isosceles trapezoid with $\overline{AB}\cong \overline{CD}.$ By the Isosceles Trapezoid Base Angles Theorem, the base angles are congruent, that is, $\angle A\cong \angle D.$

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

If the Diagonals of a Trapezoid Are Congruent, Then It Is Isosceles

For the converse, consider a trapezoid with congruent diagonals.

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

Since $\overleftrightarrow{AD}\parallel \overline{BC}$ and $\overline{BD}\parallel \overline{CP},$ $BCPD$ is a parallelogram. Therefore, $\overline{CP}\cong \overline{BD}$ and then, by the Transitive Property of Congruence, $\overline{CP}\cong \overline{AC}.$ This makes $△ACP$ an isosceles triangle.

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