A rectangle is a special kind of parallelogram, where all the angles are right angles. We have proven one property of a rectangle earlier - in a rectangle, the diagonals are equal.

Here, we will prove the converse: if a parallelogram has equal diagonals, it is a rectangle.

ABCD is a parallelogram with equal diagonals, AC=BD. Show that ABCD is a rectangle.

## Problem

## Strategy

As this is a converse theorem to the one we already proved, we will follow the strategy we often used for converse theorems - look at the way we proved the original theorem, and substitute what we needed to prove there with what was given there.

So, in the original theorem, we had shown that triangles ΔBAD and ΔCDA are congruent, using the fact that in a rectangle, m∠BAD = m∠CDA=90°. Once we had our congruent triangles, we got that the diagonals are equal, AC = BD, as corresponding parts in congruent triangles. Here, we will use the fact that we know that the diagonals are equal, AC = BD (given), to show that ΔBAD and ΔCDA are congruent. And from that , we will get that m∠BAD = m∠CDA=90°, as corresponding parts in congruent triangles

## Proof

(1) AD= AD //common side

(2) ABCD is a parallelogram //given

(3) AB= CD // (2), Opposite sides in a parallelogram are equal

(4) AC = BD //Given, diagonals are equal

(5) ΔBAD ≅ ΔCDA // Side-Side-Side postulate.

(6) ∠BAD ≅ ∠CDA // Corresponding angles in congruent triangles (CPCTC)

(7) m∠BAD = m∠CDA //(6) definition of congruent angles

(8) m∠BAD + m∠CDA =180° //(2), consecutive angles of a parallelogram are supplementary

(9) 2m∠BAD = 180° //(7), (8), substitution

(10) m∠BAD = m∠CDA = 90° //(7), (9)

(11) ABCD is a rectangle //(10), (2) defintion of a rectangle