If in a triangle $A B C$ , the altitude $A M$ be the bisector of $∠ B A D$ , where $D$ is the midpoint of side $B C$ , then prove that $(b^{2} - c^{2}) = a^{2} / 2$ .

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Solution

$∠ D A M$ = $∠ B A M$
$∠ A M D$ = $∠ A M B = 90^{o}$
and side AM (common side)
So,there is ASA congruency in triangles AMD and AMB.

Hence DM = MB

Also $B D = \frac{a}{2}$

$D M = M B =$ $\frac{B D}{2} = \frac{a}{4}$

Now using Pythagoras Theorem

$C M^{2} + A M^{2} = A C^{2}$

$\Rightarrow {(\frac{3 a}{4})}^{2} + A M^{2} = b^{2}$ ........(1)

$B M^{2} + A M^{2} = A B^{2}$

$\Rightarrow {(\frac{a}{4})}^{2} + A M^{2} = c^{2}$ ........(2)

Subtracting (2) from (1), we get

${\frac{a}{2}}^{2} = (b^{2} - c^{2})$

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