Question

In a ΔABC, angle A is twice angle B. Then,

(a) a² = b (a + c) 

(b) a² = \(\sqrt{bc}\) 

(c) a² = b (b + c) 

(d) a² = b + c

Answer 1

(c) a² = b(a+c)

In Δs ABC and DAC 

∠ABC = ∠DAC 

(∵ AD bisects ∠A and ∠A = 2∠B) 

Also, ∠ACB = ∠DCA 

⇒ DABC ~ DAC (AA similarity)

\(\frac{AC}{DC}=\frac{BC}{AC}\) ⇒ AC² = BC.DC

⇒ DC = \(\frac{AC^2}{BC} = \frac{b^2}{a}\)                    .....(i)

(In a ΔABC, AB = c, BC = a, AC = b) 

Also, AD being the angle bisector of ∠A,

\(\frac{BD}{CD}=\frac{AB}{AC}\) = \(\frac{c}{b}\)

∴ \(\frac{BD}{CD}+1\) = \(\frac{c}{b}\) + 1

⇒ \(\frac{BD + CD}{CD}=\frac{c+b}{b}\)

⇒ \(\frac{BC}{CD} =\frac{C+b}{b}\) ⇒ \(\frac{a}{CD}\) = \(\frac{c+b}{b}\) ⇒ CD = \(\frac{ab}{b+c}\)                   ......(ii)

∴ From (i) and (ii), \(\frac{b^2}{a}\) = \(\frac{ab}{b+c}\) ⇒ a² = b(a+c).