Question

In ΔABC, m∠A = 90, AD is an altitude. Therefore BD.DC = ......
(a) AB²
(d) AD²
(c) BC²
(d) AC²

Answer 1

There's a proof acc to which in a triangle ABC, if AD is perpendicular to BC then AD² = BD*DC.

So, it is quite obvious now that option (b) AD² is the correct choice. Now lets see how.

Applying pythagoras theorem we get,

AB² = AD² + BD² → 1

AC² = AD² + CD² → 2

or, adding the two we get, AB² + AC² = 2AD² + BD² + CD²

or, AB² + AC² = 2BD.DC + BD² + CD².

or, AB² + AC² = (BD + CD)²

or, AB² + AC² = BC²

Thus, it was only possible to prove angle A = 90° when AD² = BD*DC.

Answer 2

Given : In ∆ABC , m<A = 90° ,

and ,AD perpendicular to BC .

In ∆CAB and ,∆CDA ,

<C = <C [ common angle ]

<CAB = <CDA = 90°

∆CAB ~ ∆CDA ( By A.A similarity )---( 1 )

Similarly , ∆CAB ~ ∆ADB---( 2 )

from ( 1 ) and ( 2 ) , we get ,

∆CDA ~ ∆ADB

Therefore ,

DC/AD = AD/DB

=> AD² = BD × DC

Option ( b ) is correct .

I hope this helps you.

: )