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