Question

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9. Prove that any point on the bisector of an angle is equidistant from
the arms of the angle.
(Hint: Take any point P on bisector BD of ZABC. Draw PM I on AB and
PN 1 BC. Then prove APBM=APBN (by ASA criterion of congruence).
At last, prove PM = PN (as corresponding parts of congruent triangles
are equal)]
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Answer 1

Answer:

Prove that any point on the bisector of an angle is equidistant from

the arms of the angle.

(Hint: Take any point P on bisector BD of ZABC. Draw PM I on AB and

PN 1 BC. Then prove APBM=APBN (by ASA criterion of congruence).

At last, prove PM = PN (as corresponding parts of congruent triangles

are equal)]