Question
Prove that a cyclic parallelogram is a rectangle.
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Answer
Given, ABCD is a cyclic parallelogram. To prove, ABCD is rectangle.
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$\begin{array}{c}(i)(2 a-7 b)^3=(2 a)^3+(-7 b)^3+3(2 a)(-7 b)(\_\_\_\_\_\_) \\ =\_\_\_\_\_\_+\left(-343 b^3\right)+(-42 a b)(2 a-7 b) \\ =8 a^3-\_\_\_\_\_\_\_-84 a^2 b+\_\_\_\_\_\_\end{array}$
$\begin{array}{c}(ii) 95 \times 105=(100+\_\_\_\_\_\_)\left(100-\_\_\_\_\_\_\right) \\ =(100)^2-(\_\_\_\_\_\_)^2 \\ =\_\_\_\_\_\_-\_\_\_\_\_\_\\=9975\end{array}$
→$\begin{array}{c}(i)(2 a-7 b)^3=(2 a)^3+(-7 b)^3+3(2 a)(-7 b)(\_\_\_\_\_\_) \\ =\_\_\_\_\_\_+\left(-343 b^3\right)+(-42 a b)(2 a-7 b) \\ =8 a^3-\_\_\_\_\_\_\_-84 a^2 b+\_\_\_\_\_\_\end{array}$
$\begin{array}{c}(ii) 95 \times 105=(100+\_\_\_\_\_\_)\left(100-\_\_\_\_\_\_\right) \\ =(100)^2-(\_\_\_\_\_\_)^2 \\ =\_\_\_\_\_\_-\_\_\_\_\_\_\\=9975\end{array}$
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