Question

P is any point in the angle ABC such that the perpendiculars drawn from P on AB and BC are equal. Prove that BP bisects angle ABC.

Answer 1

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Answer 2

BP bisects <ABC

Explanation:

Given: P is any point in the angle ABC such that the perpendiculars drawn from P on AB and BC are equal.

PM = PN

BP is common between the two triangles PMB and PNB.

Two right-angled triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle.

So triangles PMB and PNB are congruent.

Hence corresponding angles in the two triangles will be equal.

So <ABP = <PBC

Therefore BP bisects <ABC. Hence proved.