Question

I need answer for this question​

Answer 1

Given : ABCD is a parallelogram , angle DAB =60° .AP and BP are angle bisectors of Angle A and angle B respectively .

To prove : P is mid point of CD or DP = PC

Proof : Since , AP is the bisector of angle A ,

So , angle DAP = angle DAB/2

=60°/2 =30°

We know that , Adjacent angles of parallelogram sums up to 180° .

So angle A + angle B = 180°

angle A/2 + angle B/2 = 180°/2

60°/2 + angle PBC = 90°

30° + angle PBC = 90°

angle PBC = 90° -30°

angle PBC = angle PBA = 60°

The opposite angles of parallelogram are equal , so Angle C = angle A = 60°

and angle D = angle B = 60° +60°

=120°

Since , DC // AB

angle CPB = angle PBA = 60°

In triangle ADP ,

Angle DAP + angle D + angle APD =180° (angle sum property of a triangle)

30° +120° +angle APD =180°

angle APD = 180° -30° -120°

angle APD = 30°

Now , angle APD = angle DAP = 30°

therefore , AD = DP (sides opposite to equal angles are equal) ______(1)

In triangle , BPC

angle BPC = angle PBC = Angle PCB = 60° (from earlier equations)

Hence , PBC is an equilateral triangle , so PB = BC = CP _______(2)

Now , we know that opposite sides of parallelogram are equal ,

therefore , AD = BC ______(3)

From equation 1 , 2 and 3

AD = DP = BC = PB = CP

So , DP = PC

Hence P is the mid point of DC .