Notifications
Clear all
Triangles
1
Posts
2
Users
0
Likes
415
Views
0
06/06/2021 1:07 pm
Topic starter
In Figure, D is a point on side BC of ∆ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠BAC.
1 Answer
0
06/06/2021 1:09 pm
In the given figure, let us extend BA to P such that;
AP = AC.
Now join PC.
Given, BD/CD = AB/AC
⇒ BD/CD = AP/AC
By using the converse of basic proportionality theorem, we get,
AD || PC
∠BAD = ∠APC (Corresponding angles) ……………….. (i)
And, ∠DAC = ∠ACP (Alternate interior angles) …….… (ii)
By the new figure, we have;
AP = AC
⇒ ∠APC = ∠ACP ……………………. (iii)
On comparing equations (i), (ii), and (iii), we get,
∠BAD = ∠APC
Therefore, AD is the bisector of the angle BAC.
Hence, proved.