In the given fig, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB² = BC .BD
(ii) AC² = BC. DC
(iii) AD² = BD . CD

Solution

Proof : (i) In ∆BAC and ∆BDA,
∠BAC = ∠BDA    ...(i)
In right triangle ABC, we have
∠BAC + ∠CBA = 90°    ...(ii)
In right triangle ABD, we have
∠BDA + ∠CBA = 90°    ...(iii)
Comparing (ii) and (iii), we get
∠BAC = ∠BDA
And,    ∠ACB = ∠DAB
[Each equal to 90°]
∴    ∆BAC ~ ∆BDA
[Using AA similar condition]
∴ $BA over BD equals BC over BA$
[∵ Corresponding sides of two similar triangles are proportional]
⇒    BA² = BC . BD
⇒    AB² = BC . BD.

(ii) In right triangle ACB and DCA,
∠ACB = ∠DCA = 90°
∠BAC = ∠ADC
∴    ∆ACB ~ ∆DCA
[Using AA similar condition]
∴    $AC over DC equals BC over AC$
[∵ Corresponding sides of two similar triangles are proportional]
⇒    AC² = BC × DC.

(iii) In right triangle ADB and ∆CDA,
∠DAB = ∠DCA = 90°
∠BDA = ∠ADC (common)
∴ ∆ADB ~ ∆CDA
[Using AA similar condition]
∴ $AD over CD equals BD over AD$

[∵ Corresponding sides of two similar triangles are proportional]
⇒ AD² = BD × CD.

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Triangles

In the given fig, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB² = BC .BD
(ii) AC² = BC. DC
(iii) AD² = BD . CD

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Triangles

In the given fig, ABD is a triangle right angled at A and AC ⊥ BD. Show that(i) AB2 = BC .BD(ii) AC2 = BC. DC(iii) AD2 = BD . CD

Some More Questions From Triangles Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

In the given fig, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB² = BC .BD
(ii) AC² = BC. DC
(iii) AD² = BD . CD