Question

If the diagonal of parallelogram are equal ,the show that it is rectangle.

Answer 1

Answer:

PROVED BELOW

Step-by-step explanation:

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.    

Answer 2

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Given ▶

A parallelogram PQRS ,in which PR and QS
are diagonals of parallelogram and both are equal

i.e PR = QS

TO prove :PQRS is a rectangle.

Proof ▶∆PQR and ∆ PQS

here ,

PQ = PQ (common )

PR =QS (Given)

QR=PS (opposite sides of parallelogram)

Therefore,

∆PQR ≅∆ PQS (By S.S.S congruency)

⇒ ∠PQR=∠PQS [c.p.c.t.]

Also,

∠PQR+∠PQS  = 180°(co- interior angles )

i.e ∠PQR+∠PQR=180°( because  ∠PQR=∠PQS proved above )

2 ∠PQR=180°
∠PQR= 90°

Hence , parallelogram PQRS is a rectangle .

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