Question

if diagnoal of parallelogram is equal then show that it is a rectangle ​

Answer 1

Given :-

• A parallelogram ( ||gm ) in which AC = BD.

To prove :-

• ABCD is a rectangle.

According to the question,

➞ AB = DC (Opposite sides of a parallelogram)

➞ BC = CB (Common)

➞ AC = DB (Given)

∴ Δ ABC ≅ Δ DCB (SSS - criteria)

∴ ∠ ABC = ∠ DCB ......i)

Now,

➞ DC || AB, and CB cuts them

➞ ∠ ABC + ∠ DBC = 180° (Co - interior angle)

➞ ∠ ABC = ∠ DCB = 90° [ using equation (i) ]

Thus,

ABCD is a parallelogram one of whose angle is 90°.

• So, it is proved that ABCD is a rectangle.

Answer 2

Answer:

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 (Opposite sides of a parallelogram)

⇒ ∆ ABC ≌ ∆ BAD (By SSS congruence axiom)

⇒ ∠ABC = ∠BAD (c.p.c.t)

Again,

⇒ ∠ABC + ∠BAD = 180° (co-interior angles)

⇒∠ABC + ∠ABC = 180° (∠ABC = ∠BAD)

⇒ 2∠ABC = 180°

⇒ ∠ABC = 1 ÷ 2 × 180°

⇒ ∠ABC = 90°

Hence, parallelogram ABCD is a rectangle.