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Q2 In Fig. 6.57, D is a point on hypotenuse AC of D ABC, such that BD $\perp$ AC, DM $\perp$ BC and DN $\perp$ AB. Prove that : $DN^2 = DM . AN$

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In $\triangle$ DBN,

$\angle 5+\angle 7=90 \degree.......................1$

In $\triangle$ DAN,

$\angle 6+\angle 8=90 \degree.......................2$

BD $\perp$ AC, $\therefore \angle ADB=90 \degree$

$\angle 5+\angle 6=90 \degree.......................3$

From equation 1 and 3, we get $\angle 6=\angle 7$

From equation 2 and 3, we get $\angle 5=\angle 8$

In $\triangle DNA\, \, and\, \, \triangle BND,$

$\angle 6=\angle 7$

$\angle 5=\angle 8$

$\triangle DNA\, \, \sim \, \, \triangle BND$ (By AA)

$\Rightarrow \frac{AN}{DN}=\frac{DN}{NB}$

$\Rightarrow \frac{AN}{DN}=\frac{DN}{DM}$ (NB=DM)

$\Rightarrow$ $DN^2 = AN . DM$

Hence proved

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