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

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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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)

\RightarrowDM^2 = DN . MC

Hence proved 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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