Question

AX is the bisector of angle BAC; P is any point on AX. prove that the perpendiculars drawn from P to AB and AC are equal.

Answer 1

Answer:

The proof is explained in Step-by-step explanation

Step-by-step explanation:

Given AX is the bisector of angle BAC and P is any point on AX we have to prove that the perpendiculars drawn from P to AB and AC are equal.

In ΔPAQ and ΔPAR

∠PAQ = ∠PAR         (∵Given )

∠PQA = ∠PRA        ( ∵ each 90°)

AP = AP                   ( ∵ common )

By AAS rule, ΔPAQ ≅ ΔPAR

Therefore, by CPCT i.e by Corresponding parts of Congruent triangles PQ = PR

Hence, the perpendiculars drawn from P to AB and AC are equal.

Answer 2

Answer:

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