Question

hello brainlians!!
Question is in the attachment.

Solve Question no. 2 ​

Answer 1

Answer:

Step-by-step explanation:

Given: 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  

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 ...(2) _ Sum of consecutive interior angles on the same side of the transversal is  

180  

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From (1) and (2) ,

∠BAD=∠ABC=90  

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∴∠A=90  

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 and ∠C=90  

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Parallelogram ABCD is a rectangle.

Answer 2

$\bf \: ∠A+∠A=180°$

$\bf \: 2∠A=180°$

$\bf \: ∠A= \cancel\frac{180 \degree}{2}$

$\bf \: ∠A=90°$

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It helps you!

@ShiningBlossom