Question

D is a point on side bc of ABC such that bd/cd=ab/ac. Prove that ad is the bisector of angle bac

Answer 1

GIVEN: BD/DC = AB/AC

TO PROVE: < BAD = < CAD

CONSTRUCTION: Extend BA to E, such that AE = AC .

PROOF: Since AE = AC

Hence < AEC = < ACE …

Since, BD/ CD = AB/AC ( given)

But , AC = AE

=> BD/CD= AB/ AE

=> AD // EC ( As intercepts on transversals BC & BE are proportional..)

So, < CEA = < DAB ( consecutive interior angles formed by // lines) …… (1)

< ECA = < CAD (alternate interior angles...AD// EC ) …….. (2)

But < CEA = < ECA

Therefore, < DAB= < CAD ( bu 1 & 2 )

Answer 2

Answer:

Given: ABC is a triangle and D is a point on BC such that

To prove: AD is the internal bisector of BAC.

Construction: Produce BA to E such that AE = AC. Join CE.

Proof: In AEC, since AE = AC

AEC = ACE ……….(i)

[Angles opposite to equal side of a triangle are equal]

Now, [Given]

[ AE = AC, by construction]

By converse of Basic Proportionality Theorem,

DA CE

Now, since CA is a transversal,

BAD = AEC ……….(ii) [Corresponding s]

And DAC = ACE ……….(iii) [Alternate s]

Also AEC = ACE [From eq. (i)]

Hence, BAD = DAC [From eq. (ii) and (iii)]

Thus, AD bisects BAC internally.