Line
is the bisector of an
and B is any point on
. BP and BQ are perpendiculars from B to the arms of
(See figure). Show that:
(i)
(ii) BP=BQ or B is equidistant from the arms of
.
Solution (i)
In
and
, we have
(It is given that AB is the bisector) (1)
(Common) (2)
(Each equal to
) (3)
Therefore, by AAS congruence rule, we have
.
Solution (ii)
From solution (i), we know that
.
Therefore, BP=BQ (Corresponding parts of congruent triangles are equal)
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