In the fig. 3.76 the point P is in the interior of $∆$ ABC, such that ray BP is the angle bisector of $∠$ ABC but ray CP is not an angle bisector of $∠$ ACB. Then prove that the ray AP is not angle bisector of $∠$ BAC.

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Solution

$Given : BP is the angle bisector of ∠ ABC, but CP is not the angle bisector of ∠ ACB . Thus, by angle - bisector theorem, P is equidistant from AB and AC . Or, PX = PZ . . . (i) A nd P is not equidistant from AC and BC . Or, PZ \neq PY . . . . . . (ii) From (i) and (ii), we get : PX \neq PY Or, P is not equidistant from AB and AC . Or, AP is not an angle bisector of ∠ BAC . Hence p roved .$

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In the fig. 3.76 the point P is in the interior of ∆ABC, such that ray BP is the angle bisector of ∠ABC but ray CP is not an angle bisector of ∠ACB. Then prove that the ray AP is not angle bisector of ∠BAC.

In the fig. 3.76 the point P is in the interior of $∆$ ABC, such that ray BP is the angle bisector of $∠$ ABC but ray CP is not an angle bisector of $∠$ ACB. Then prove that the ray AP is not angle bisector of $∠$ BAC.