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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.
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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