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Q:D is the midpoint of the side BC of triangle ABC, If P and Q are points on AB and AC such that DP bisects angle BDA and DQ bisects angle ADC , then prove that PQ is parallel to BC.



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The angular bisector of angle in a triangle divides the opposite segment, in the same ratio as the other two sides.

Since  DP bisects angle ADB. =>  BP/AP  = BD / AD
Add 1 to both sides:    (BP+AP)/AP = BD/AD + 1
   =>  AB/AP = 1 + BD/AD
Since  DQ bisects angle ADC,  =>  CQ /AQ = CD/ AD
   => (CQ + AQ)/AQ = 1 + CD/AD
  =>  AC/AQ = 1 + CD/AD

since CD = BD,  D is the midpoint of BC,
     AB/AP  =  AC / AQ        or  AB/ AC = AP / AQ

AB is parallel to AP  and AC is parallel to AQ.  And corresponding ratios are same.  The angle A is common in the two triangles ABC and APQ.

Hence, the two triangles  ABC and  APQ are similar.

Hence, PQ is parallel to BC. 

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