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

Q2 (2)     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

Answers (1)
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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 \perpAC, \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)

\RightarrowDN^2 = AN . DM

Hence proved 

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