Question

Prove that the external bisector of an angle of a triangle divides the opposite side externally n the ratio of the sides containing the angle.

Answer 1

In ΔABC, CE || AD

∴ `"BD"/"CD" = "AB"/"AE"`.....(i)

(By Basic Proportionality theorem)
AD is e bisector of ∠CAF
∠FAD = ∠CAD......(ii)
Since CE || AD
Therefore,
∠ACE = ∠CAD......(iii)  ...(alternate angles)
∠AEC = ∠FAD......(iv)   ...(corresponding angles)
From (ii) and (iii) and (iv)
∠AEC = ∠ACE
In ΔAEC,
∠AEC = ∠ACE
AC = AE  ......(v) ...(Equal angles have equal sides opposite to them)
From (i) and (v)

`"BD"/"CD" = "AB"/"AC"`.