exam

Asked by  | 25th 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  | 9th Aug, 2009, 12:17: AM

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