Question

if the diagonals of a parallelogram are equal then show that it is a rectangle​

Answer 1

Answer:-

★ Given:-

AC = BD

★ To show that:-

ABCD is a rectangle if the diagonals of a parallelogram are equal

★ To prove that:-

One of its interior angle is right angled.

★ Proof:-

In ΔABC and ΔBAD,

BC = BA (Common)

AC = AD (Opposite sides of a parallelogram are equal)

AC = BD (Given)

Therefore, ΔABC ≅ ΔBAD [SSS congruency]

∠A = ∠B [CPCT]

★ Also,

∠A + ∠B = 180° (Sum of the angles on the same side of the transversal)

→ 2∠A = 180°

→ ∠A = 90° = ∠B

Therefore, ABCD is a rectangle.

Hence Proved.

Answer 2

Given :-

• A parallelogram ABCD , in which AC = BD

TO Prove : -

• ABCD is a rectangle .

Proof : -

● In △ABC and △ABD

→ AB = AB [common]

→ AC = BD [given]

→ BC = AD [opp . sides of a | | gm]

→ △ABC ≅ △BAD [ by SSS congruence axiom]

→ ∠ABC = △BAD [c.p.c.t.]

Also, ∠ABC + ∠BAD = 180° [co - interior angles]

→ ∠ABC + ∠ABC = 180° [∵ ∠ABC = ∠BAD]

→ 2∠ABC = 180°

→ ∠ABC = 1 /2 × 180° = 90°

Hence !!

parallelogram ABCD is a rectangle.