Given: A parallelogram ABCD in which AC = BD.
To prove: Parallelogram ABCD is a rectangle
Proof: In ΔBAD and ΔABC,
AB = BA (common sides)
BD = AC (diagonals of a parallelogram are equal)
AD = BC (opposite sides of a parallelogram are equal)
ΔBAD = ΔABC (by SSS congruency rule)
⇒ ∠BAD = ∠ABC (by c.p.c.t)
But ∠BAD + ∠ABC = 180° (sum of interior angles on the same side of a transversal is 180°)
⇒ 2 ∠BAD = 180° [∵ ∠BAD = ∠ABC]
⇒ ∠BAD = 90°
Hence, parallelogram ABCD is a rectangle.
