CBSE Class 9 Answered
exam
Asked by | 25 Jul, 2009, 05:26: AM
Expert Answer
consider the angle BAC and its internal bisector AD.
Take any point P on AD.
Drop perpendiculars PE and PF on BA and AC respectively.
in right triangles,
EAPand FAP
AP=AP....common side
angle EAP=angle FAP...AD is bisector given
angle APE= angle APF...90 degrees each
so the triangles are congruent by RHS rule.
so
PE=PF(C.P.C.T)
Answered by | 09 Aug, 2009, 12:17: AM
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