If the diagonal of parallelogram are equal then prove that it is a square
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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
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
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halimashadiya123:
I had not asked to prove rectangle.
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Given: ABCD is a parallelogram
AO=OD
CO=OB
To Prove: ABCD is a parallelogram
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