Question

In ∆ABC , angle B = 90° and seg BD perpendicular to side AC , A-D-C, then ............... plz give me answer fast ..... itcs ergent ​

Answer 1

Answer:

please send the image of the figure with the question.

Answer 2

This can be proved in various ways. Let me try one,

Refer following figure,

Refer triangle, $\triangle ABD$

$\tan \theta=\displaystyle \dfrac{AD}{BD} \tag 1$

Refer triangle [math]\triangle ACD[/math],

$\tan (90-\theta)=\displaystyle \dfrac{AD}{CD}$

$\cot \theta=\displaystyle \dfrac{AD}{CD} \tag 2$

Multiplying $(1), (2)\implies$

$\tan \theta ×\cot \theta=\displaystyle \dfrac{AD}{BD}×\dfrac{AD}{CD} =1$

$\boxed{AD^2=BD × CD}\, \blacksquare$

Refer following figure

In $\triangle ABC,$ if $\angle ABC=\theta$, $\angle ACB=(90-\theta)$

$\therefore$ in $\triangle ACD, \angle CAD=90-(90-\theta)=\theta$

Refer triangle, $\triangle ABD$

[math]\tan \theta=\displaystyle \dfrac{AD}{BD} \tag 1[/math]

Refer $\triangle ACD$,

$\tan \theta=\displaystyle \dfrac{CD}{AD} \tag 2$

Equating $(1), (2)\implies$

$\tan \theta =\displaystyle \dfrac{AD}{BD}=\dfrac{CD}{AD}$

$\boxed{AD^2=BD × CD}\, \blacksquare$