★ 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!!
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★ 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!!
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Verified answer
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:
[tex]\colorbox{red}{\boxed{\color{lime} \mathbb {\: CORRECT \: \: OPTION \: \: IS \: \: (A).}}}[/tex]
Step-by-step explanation:
[tex] \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}[/tex]
[tex] \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} \] [/tex]
[tex] \color{darkcyan}\[ \begin{array}{l} \therefore \quad \tt \triangle D A C \sim \triangle \: D B A \quad \text { \tt [By AA Similarity] } \\ \\ \tt \Rightarrow \dfrac{A D}{B D}=\dfrac{C D}{A D} \quad \text { \tt [Corresponding sides of similar triangles are proportional] } \\ \\ \tt\Rightarrow A D^{\circ}=C D \times B D \end{array} \][/tex]