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

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Join BD

Given : D is a point on hypotenuse AC of D ABC, such that BD $\perp$ AC, DM $\perp$ BC and DN $\perp$ AB.Also DN || BC, DM||NB

$\angle CDB=90 \degree$

$\Rightarrow \angle 2+\angle 3=90 \degree.............................1$

In $\triangle$ CDM, $\angle 1+\angle 2+\angle DMC=180 \degree$

$\angle 1+\angle 2=90 \degree.......................2$

In $\triangle$ DMB, $\angle 3+\angle 4+\angle DMB=180 \degree$

$\angle 3+\angle 4=90 \degree.......................3$

From equation 1 and 2, we get $\angle 1=\angle 3$

From equation 1 and 3, we get $\angle 2=\angle 4$

In $\triangle DCM\, \, and\, \, \triangle BDM,$

$\angle 1=\angle 3$

$\angle 2=\angle 4$

$\triangle DCM\, \, \sim \, \, \triangle BDM,$ (By AA)

$\Rightarrow \frac{BM}{DM}=\frac{DM}{MC}$

$\Rightarrow \frac{DN}{DM}=\frac{DM}{MC}$ (BM=DN)

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

Hence proved

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