The incircle of an isosceles triangle ABC,CB=CA, touches the sides AB,BC and CA at D,E and F respectively. Prove that D bisects AB.
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Solution
Given : CB=CA ---(1) CE=CF (Tangents from point C) ---(2) On subtracting equation (1) and (2), we get CB−CE=CA−CF BE=FA ---(3) BE=BD BD=FA ----(4) FA=AD ----(5) Therefore, from equation (4) and (5), BD=AD Hence, D bisects AB.