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In fig., a circle is inscribed in triangle ABC touches its sides AB, BC and ACat points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10cm, then find the length of AD, BE and CF.
Given: AB=12 cm, BC= 8 cm, and AC= 10 cm.Let, AD= AF = x cm, BD = BE = y cm, and CE=CF= z cm.(Tangents drwan from an external point to the circle are equal in length)⇒ 2(x+y+z )= AB+BC+AC=AD+ DB+BE+EC+AF+FC= 30 cm.⇒ x+y+z=15 cm. AB= AD+DB =x+y= 12 cm∴ z= CF =15-12 =3 cm AC=AF+FC= x+z = 10 cm∴ y= BE= 15-10 = 5 cm∴ x=AD=x+y+z-z-y= 15-3-5=7 cm.
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