ABCD IS A PARALEELOGRAM AND ANGLE DAB =60 . IF THE BISECTORS AP AND BP OF ANGLES A AND B RESPECTIVELY MEET AT P ON CD . PROVE THAT P IS MID PT OF CD
Asked by ayush kumar
| 26th Nov, 2010,
03:05: PM
Expert Answer:
Dear student,
PAB=
PAD= 30° ….as AP is bisector
APD = 30°…..as AB ll CD
In triangle APD,
AD=DP………(i) ….sides opposite to equal angles
Similarly, in triangle PBC
PBC =
BPC= 60°
BC=PC…..(ii) ..sides opposite to equal angles
CP=PD….from (i) and (ii) as AD=BC being opp sides of a parallelogram
Thus, P is the mid point of CD.
We hope that clarifies your query.
Regards,
Team
TopperLearning
PAB=
PAD= 30° ….as AP is bisector
APD = 30°…..as AB ll CD
In triangle APD,
AD=DP………(i) ….sides opposite to equal angles
Similarly, in triangle PBC
PBC =
BPC= 60°
BC=PC…..(ii) ..sides opposite to equal angles
CP=PD….from (i) and (ii) as AD=BC being opp sides of a parallelogram
Thus, P is the mid point of CD.
We hope that clarifies your query.
Regards,
Team
TopperLearning
Answered by
| 27th Nov, 2010,
07:20: PM
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