Question

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Question for Math experts! The perpendicular from A on the side BC of a triangle ABC intersects BC at D such that BD : CD = 3:2. Prove that AB^2 - AC^2 = 1/5 BC^2

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Answer 1

Diagram in attachment :)

Given :

BD : CD = 3 : 2

By Pythagoras theorem :

AB² = AD² + BD²

= > AD² = AB² - BD²

By Pythagoras Theorem :

AC² = AD² + CD²

= > AD² = AC² - CD²

From ( 1 ) and ( 2 ) :

AB² - BD² = AC² - CD²

= > AB² - AC² = BD² - CD²

But BD / CD = 3 / 2

= > BD = 3/2 CD

Insert this in the equation :

AB² - AC² = ( 3/2 CD )² - CD²

= > AB² - AC² = 9/4 CD² - CD²

= > AB² - AC² = ( 9 - 4 CD² ) / 4

= > AB² - AC² = 5/4 CD²

Observe the following :

BC² = ( BD + CD )²

= > BC² = ( 3/2 CD + CD )²

= > BC² = ( 5/2 CD )²

= > BC² = 25/4 CD²

= > CD² = 4/25 BC²

Put this in the equation :

= > AB² - BD² = 5/4 × 4/25 BC²

= > AB² - BD² = 1/5 BC²

Hence Proved !! It was fun to solve ^^"

Answer 2

QUESTION :

The perpendicular from A on the side BC of a triangle ABC intersects BC at D such that BD : CD = 3:2. Prove that AB^2 - AC^2 = 1/5 BC^2

ANSWER :

GIVEN :

In ∆ABC

AD is perpendicular to BC

BD:CD=3:2

so BD/CD = 3/2

So BD = 3CD/2......1

adding CD on both sides

BD+CD=(3CD/2) + CD

BC= (3CD+2CD)/2

BC= (5CD/2)

CD = 2BC/5......2

TO PROVE :

AB²-AC²=BC²/5

PROOF :

Now in ∆ADC , Angle D =90°

so by Pythagoras theorem

we get

AC²=AD²+DC²

So AD² = AC² - CD²............3

Now similarly in ∆ADB

AB²=BC²+AD²

So AD²=AB²-BD²..............4

Now as LHS of Both sides are equal

by equating 3 and 4

we get

AC²-CD²=AB²-BD²

AB²-AC²=BD²-CD²

But from 1

AB²-AC²= (3CD/2)² -CD²

AB²-AC²= (9CD²/4) -CD²

AB²-AC²= (9CD²-4CD²)/4

AB²-AC²=5CD²/4

now from 2

AB²-AC²=(5/4) × (2BC/5)²

AB²-AC²=(5/4) × (4BC²/25)

AB²-AC²=BC²/5

hence proved

NOTE :

while solving such questions

try to find some link between the given things

and make the equation in the forms which we have to prove

also the diagram drawn must be correct