Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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)

best_answer

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

Posted by

seema garhwal

View full answer

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 10 board exams from February 15 to March 18; download date sheet 2025 PDF

anam-khan

CBSE Date Sheet 2025 Live: CBSE Class 10, 12 board exam timetable soon at cbse.gov.in; practical dates

anam-khan

CBSE Date Sheet 2025 Live: CBSE to announce Class 10, 12 board exam dates soon at cbse.gov.in; sample paper

Latest Question