Question

In a triangle ABC, AD is the angle bisector of ∠BAC and ∠BAD = 60º. What is the length of AD?

(a) \(\frac{b+c}{bc}\)

(b) \(\frac{bc}{b+c}\)

(c) \(\sqrt{b^2+c^2}\)

(d) \(\frac{(b+c)^2}{bc}\)

Answer 1

(b) \(\frac{bc}{b+c}\)

Let AD = p 

Area of ΔABC = \(\frac{1}{2}\) bc sin ∠BAC = \(\frac{1}{2}\) bc sin 120º

= \(\frac{1}{2}\) bc x \(\frac{\sqrt3}{2}\) = \(\frac{\sqrt3}{2}\)bc

Area of ΔBAD = \(\frac{1}{2}\) cp sin 60º

= \(\frac{\sqrt3}{4}\) cp

Area of ΔCAD = \(\frac{1}{2}\) bp sin 60º = \(\frac{\sqrt3}{4}\) bp

Now Area (ΔABC) = Area (ΔBAD) + Area (ΔCAD)

⇒ \(\frac{\sqrt3}{4}\) bc = \(\frac{\sqrt3}{4}\) cp + \(\frac{\sqrt3}{4}\) bp ⇒ bc = p(b + c) ⇒ p = \(\frac{bc}{b+c}\).