Question

★ Question :-

△ABD is right angled at A and AC _I_ BD then AD² = ______

(a) BD × CD

(b) BD × BC

(c) BD × CA

(d) BC × CD

↦Explanation needed!! ​

Answer 1

Option (a)

Step-by-step explanation:

Given :-

∆ ABD is a right angled triangle .

Right angle is at A.

AC ⊥ BD

To find :-

Value of AD²

Solution :-

In ∆ABD , ∠A = 90° and AC ⊥ BD

and

∆ CAD is a right angled triangle

In ∆ ABD and ∆ CAD

∠BAD = ∠ACD = 90°

∠D = ∠D ( Common side )

By A.A similarity ,

∆ ABD ~ ∆ CAD

Hence, AB / AC = BD / AD = AD / CD

Since, Ratios of corresponding sides of similar triangles are equal.

=> BD / AD = AD / CD

On applying cross multiplication then

=> AD × AD = BD × CD

=> AD² = BD × CD

Therefore, AD² = BD × CD

Answer:-

In ∆ABD , ∠A = 90° , If AC ⊥ BD then AD² = BD × CD

Used Property :-

→ AA criteria for similarity .

" If two angles of one triangle are respectively equal to two angles of another triangle then the two triangles are similar.

Answer 2

Answer:

$\colorbox{red}{\boxed{\color{lime} \mathbb {\: CORRECT \: \: OPTION \: \: IS \: \: (A).}}}$

Step-by-step explanation:

$\color{darkred}\begin{array}{ll}\text { Given } & : \triangle \tt A B D, \angle A=90^{\circ} \text { and } \tt \: A C \perp B D \\ \text { To Prove } & \text {: } \tt A D^{2}=B D . D C\end{array}$

$\color{navy}\[ \begin{array}{ll} \tt \text{ \tt \: In \( \tt\triangle D A C \) and \( \tt\triangle D B A \),} \\ \\ \angle 4=\angle 4 & \text { \tt [Common] } \\ \\ \tt \angle 3=\angle B A D & \: {\left [ \tt90^{\circ} \text { \tt \: each }\right]} \end{array} \]$

$\color{darkcyan}\[ \begin{array}{l} \therefore \quad \tt \triangle D A C \sim \triangle \: D B A \quad \text { \tt [By AA Similarity] } \\ \\ \tt \Rightarrow \dfrac{A D}{B D}=\dfrac{C D}{A D} \quad \text { \tt [Corresponding sides of similar triangles are proportional] } \\ \\ \tt\Rightarrow A D^{\circ}=C D \times B D \end{array} \]$