If the diagonals of a parallelogram are equal then show that it is a rectangle
Answers
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.
Step-by-step explanation:
Gven: In parallelogram ABCD, AC=BD
To prove : Parallelogram ABCD is rectangle.
Proof : in △ACB and △BDA
AC=BD ∣ Given
AB=BA ∣ Common
BC=AD ∣ Opposite sides of the parallelogram ABCD
△ACB ≅△BDA∣SSS Rule
∴∠ABC=∠BAD...(1) CPCT
Again AD ∥ ∣ Opposite sides of parallelogram ABCD
AD ∥BC and the traversal AB intersects them.
∴∠BAD+∠ABC=180∘
...(2) _ Sum of consecutive interior angles on the same side of the transversal is
180∘
From (1) and (2) ,
∠BAD=∠ABC=90∘
∴∠A=90∘
and ∠C=90∘
Parallelogram ABCD is a rectangle.
and ∠C=90∘
Parallelogram ABCD is a rectangle.
