In the figure given below AD is bisector of ∠BAC. If AB=6cm,AC=4cm and BD=3cm, find BC.
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Solution
From the question it is given that, AD is bisector of ∠BAC AB=6cm,AC=4cm and BD=3cm Construction, from C draw a straight line CE parallel to DA and join AE ∠1=∠2 … [equation (i)] By construction CE∥DE So, ∠2=∠4 … [because alternate angles are equal] [equation (ii)] Again by construction CE∥DE ∠1=∠3 … [because corresponding angles are equal] [equation (iii)] By comparing equation (i), equation (ii) and equation(iii) we get, ∠3=∠4 So, AC=AE … [equation (iv)] Now, consider the △BCE, CE∥DE BD/DC=AB/AE BD/DC=AB/AC 3/DC=6/4 By cross multiplication we get, 3×4=6×DC DC=(3×4)/6 DC=12/6 DC=2 Therefore, BC=BD+DC =3+2 =5cm.