Question

3. Prove that parallelogram is a rectangle if its diagonals are of equal lengths.​

Answer 1

Answer:

Let's draw a figure (See file attached below, I used ms paint)

Remember, I drew a rectangle because we have to prove that it is an rectangle.

Given: ABCD is a parallelogram with AC=BD.

Now,

In △ABC and △ABD

• AB = AB         [common side]

• AC = BD              [given]

• BC = AD       [opposite sides of a║gm are equal.]

⇒ △ABC ≅ △BAD [ by SSS congruence]

Then, by CPCT, ∠ABC = △BAD

Now, ∠ABC + ∠BAD = 180°----(2)  [interior angles on the same side of the transversal are sypplementary]

But, ∠ABC = △BAD -----(1)

From (1) and (2), we get:

∠ABC + ∠ABC = 180° [∵ ∠ABC = ∠BAD]

⇒ 2∠ABC = 180°  

⇒ ∠ABC = 180°÷2

⇒ ∠ABC = 90°  

Now, we know that If one of the interior angles of a parallelogram is 90°, is is an rectangle.

Since ABCD is a parallelogram and one of its interior angles is 90º, ABCD is a rectangle.

HOPE THIS HELPS :D