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Question

# In a ΔABC, AD is the bisector of ∠A. (i) If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC. (ii) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC. (iii) If AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm, find AC. (iv) If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find BC.

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Solution

## (i) AB = 6.4 cm, AC = 8 cm, BD = 5.6 cm Let BC = x Now, DC = (BC-BD) = (x-5.6) cm In ΔABC, AD is the base of ∠A So, by the angle bisector theorem, we have BD/DC=AB/AC ⇒5.6/x−5.6=6.4/8 ⇒⇒ 6.4x - 35.84 = 44.8 ⇒⇒ 6.4x = 80.64 ⇒⇒ x = 12.6 Hence, BC = 12.6 cm and DC = (12.6 - 5.6) cm = 7 cm (ii) AB = 10 cm, AC = 14 cm, BC = 6 cm Let BD = x, DC = (BC - BD) = (6-x) cm In ΔABC, AD is the bisector of ∠A So, By angle bisector theorem, We have, BD/DC=AB/AC ⇒ x/6−x=10/14 ⇒ 14x = 60 - 10x ⇒ 24x = 60 x = 2.5 Hence, BD = 2.5 cm and DC = (6 - 2.5) cm = 3.5 cm (iii) AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm DC = BC - BD = (6-3.2) cm = 2.8 cm Let, AC = x, In ΔABC, AD is the bisector of ∠A So, by the angle bisector theorem we have Therefore, BD/DC=AB/AC ⇒ 3.2/2.8=5.6/x ⇒ x= 5.6×2.8/3.2=4.95cm Hence, AC = 4.9 cm (iv) AB = 5.6 cm, AC = 4 cm, DC = 3 cm Let BD = x, In ΔABC, AD is the bisector of ∠A So, by the angle bisector theorem we have Therefore, BD/DC=AB/AC ⇒ x/3=5.6/4 ⇒ x= 5.6×3/4=4.2 cm Hence, BD = 4.2 cm So BC = BD + AC = (4.2 + 3) cm BC = 7.2 cm

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