In $△ M G N$ , $M P ⊥ G N$ . If $M G = a$ units, $M N = b$ units, $G P = c$ units and $P N = d$ units.
Prove that $(a + b) (a - b) = (c + d) (c - d)$ .

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Solution

In $△ M P G$ , $∠ P = 90^{0}$ , $M G = a$ units and $G P = c$ units.

Using pythagoras theorem, we have

$M G^{2} = G P^{2} + M P^{2} \Rightarrow a^{2} = c^{2} + M P^{2} \Rightarrow M P^{2} = a^{2} - c^{2} . . . . . . (1)$

In $△ M P N$ , $∠ P = 90^{0}$ , $M N = b$ units and $P N = d$ units.

Again using pythagoras theorem, we have

$M N^{2} = M P^{2} + P N^{2} \Rightarrow b^{2} = M P^{2} + d^{2} \Rightarrow M P^{2} = b^{2} - d^{2} . . . . . . (2)$

Equate equations 1 and 2 as follows:

$a^{2} - c^{2} = b^{2} - d^{2} \Rightarrow a^{2} - b^{2} = c^{2} - d^{2} \Rightarrow (a - b) (a + b) = (c - d) (c + d) (∵ a^{2} - b^{2} = (a - b) (a + b))$

Hence, $(a - b) (a + b) = (c - d) (c + d)$ .

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