Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Q3 In Fig. 6.53, ABD is a triangle right angled at A and AC $\perp$ BD. Show that $AC^2 = BC . DC .$

Answers (1)

best_answer

Let $\angle CAB$ be x

In $\triangle ABC$ ,

$\angle CBA=180 \degree-90 \degree-x$

$\angle CBA=90 \degree-x$

Similarly,

In $\triangle CAD$ ,

$\angle CAD=90 \degree-\angle CAB$

$\angle CAD=90 \degree-x$

$\angle CDA=180 \degree-90 \degree-(90 \degree-x)=x$

In $\triangle ABC\, and\, \triangle ACD,$

$\angle CBA\, =\angle CAD$

$\angle CAB\, =\angle CDA$

$\angle ACB\, =\angle DCA$ ( Each right angle)

$\triangle ABC\, \sim \triangle ,ACD$ (By AAA)

$\frac{AC}{DC}=\frac{BC}{AC}$

$\Rightarrow AC^2=BC\times DC$

Hence proved

Posted by

seema garhwal

View full answer

Similar Questions

Trending Articles/News

anam-khan

NEP 2020 gives Indian schools space for new way of learning, with creativity, contemplation

anam-khan

CBSE Class 10 board exams 2026 for sports, olympiad students in second phase only

anam-khan

‘Some don’t even have Aadhaar’: CBSE’s APAAR ID mandate sends students scrambling before board exams

Latest Question