In a ΔABC, AD is the internal bisector of ∠A, meeting BC at D.
If AB = x, AC = x–2, BD = x + 2 and DC = x–1 find the value of x.
Given: A ΔABC with AD as internal bisector of ∠A, meeting BC at D. and AB = x, AC = x–2, BD = x + 2, DC = x–1
Required: The length of BC
Here, In ΔABC AD is the internal bisector of ∠A
∴ By angle bisector theorem
⇒
⇒ (x + 2)(x—2) = x(x—1)
⇒ x2—4 = x2—x (∵ (a + b)(a—b) = a2—b2)
⇒ x = 4
∴The value of x = 4cm