Question

hlo friends.
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please help me to do this question of class 9th plz

Answer 1

Here's your answer..

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Solution:-
We have a parallelogram ABCD in which AC=BD

We have to prove that it is an Rectangle .

So,
In∆ABC and ∆BAD
=>AC=BD(Given)
=>AB=BA(Common)
=>BC=AD(Opposite side is || are equal)

By criteria (SSS),
∆ABC is congruent ∆BAD
Hence,
= <ABC=<BAD (By c.p.c.t)

And also,
= <ABC +<BAD =180°(Co-interior angle)
= <ABC +<ABC=180°
(Since,<ABC =<BAD so we can write <ABC at place of <BAD)

=<ABC +<ABC=180
=2<ABC =180°
=<ABC = 90°

Hence,
Parallelogram ABCD is also a Rectangle.

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Hope it helps you...

Answer 2

Answer:

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