Question

In a right angled ABC, angle A = 90°, and AD*BC, prove that AD2 = BD×CD solve by using similarity​

Answer 1

First-Method:-

ABC is a right angled triangle in which angle A=90° and AD is perpendicular

to BC. Thus , In right angled triangle ABC , AB^2+AC^2=BC^2……………(1)

In right angled triangle ADB

AD^2+DB^2=AB^2……………….(2)

In right angled triangle ADC

AD^2+DC^2=AC^2……………….(3)

by adding the eqn. (2) and (3)

2.AD^2+DB^2+DC^2= AB^2+AC^2. , putting AB^2+AC^2=BC^2 from eqn. (1)

2.AD^2+DB^2+DC^2=BC^2. , putting BC=BD+CD

2.AD^2+DB^2+DC^2=(BD+CD)^2

or. 2.AD^2+DB^2+DC^2=BD^2+CD^2+2×BD×CD.

or. 2.AD^2 = 2×BD×CD.

or. AD^2= BD×CD. Proved.

Second -Method:-

In right angled triangle ABC let angle ABC= B then angle ACB= 90°-B .

Thus , in right angled triangle ADC , angle CAD = B.

In right angled triangle ADB. , tanB=AD/BD………………(1)

In right angled triangle CDA. , tanB=CD/AD………………..(2)

From eqn. (1) and (2)

AD/BD= CD/AD

or. AD^2=BD×CD. Proved.

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