Question

Prove that any cyclic parallelogram is rectangle

Answer 1

$given...abcd \: is \: a \: cyclic \: parallelogram \\ prove.. \\ abcd \: is \: a \: rectanle \\ proof.. \\ angle \: a + angle \: c = 180.. \: by \: \\ cyclic \: property \\ 2angle \: a = 180 \: by \: parallelogram \\ property \\ so \: angle \: a = 90 \: degree \\ so \: abcd \: is \: rectangle$

Answer 2

Hello mate ☺

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Solution:

➡It is given that parallelogram ABCD is cyclic. We need to prove that ABCD is a rectangle.

∠B=∠D   (Opposite angles of a parallelogram are equal) ....(1)

∠B+∠D=180°   ...... (2)  

(Sum of opposite angles of a cyclic quadrilateral is equal to 180°)                    

Using equation (1) in equation (2), we get

∠B+∠B=180°

⇒2∠B=180°

⇒∠B=180/2=90°      …...(3)

➡Therefore, ABCD is a parallelogram with ∠B=90° which means that ABCD is a rectangle.

I hope, this will help you.☺

Thank you______❤

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