Question

1. ABCD is a parallelogram. If the bisectors DP and CP of angles D and C meet at P on side AB, then show that P is the mid-point of side AB.

Answer 1

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Answer 2

Answer :

It is proved that the point P is the mid-point of the side AB.

Given ABCD a parallelogram and DP & CP be the bisectors of the angles D and C which meet at a point P on the side AB.

It has to be proved that P is the mid-point of the side AB

Now, using the properties of the parallelogram

AB = CD and AD = BC

also, A = C and B = D

as, CP is the angle bisector of the angle C then, if A = x

then BCP = x/2

and similarly DP is the angle bisector of the angle D then, B = 180° - x

then ADP = (180° - x)/2

By ASA congruence rule, both the triangles

APD ≅ BPC which implies AP = BP

Hence, P is the mid-point of the side AB.

So, to conclude in a sentence, it is proved that the point P is the mid-point of the side AB.

To know more about Parallelogram click the link below

https://brainly.in/question/6529938

To know more about Congruent Triangles click the link below

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