Question

In the given figure AB is parallel to CD. PA and PC are bisector of angle BAC and angle ACD. Find angle APC.

Answer 1

Answer:

Angle APC = 90

Step-by-step explanation:

Since AB║CD

Angle BAC = Angle DCA ( Co- Interior Angles )

Let Angle BAC be x

Since Angle BAC = Angle DCA, Angle DCA is also x

Sum of Co- Interior Angle is 180

Therefore, 2x = 180

                    x = 90

Since the two angles are bisected,

1/2 of Angle BAC = Angle PAC

1/2 * 90 = 45

1/2 of Angle DCA = Angle PCA

1/2 * 90 = 45

Angle PAC and Angle PCA are 45° each

Using Angle Sum Property of a Triangle

Angle PCA + Angle PAC + Angle APC = 180

45 + 45 + Angle APC = 180

Angle APC = 180

Therefore Angle APC = 90

Answer 2

Answer:

AB || CD

PA is the bisector of angle BAC

=> angle PAB = angle PAC

PC is the bisector of angle DCA

=> angle PCD = angle PCA

A/Q , angle BAC + angle DCA = 180°

( the sum of 2 corresponding angles is 180° => angle PAB + angle PAC + angle PCD + angle PCA = 180°

=> angle PAC + angle PAC + angle PCA + angle PCA = 180°

=> 2 angle PAC + 2 angle PCA = 180°

=> 2 ( angle PAC + angle PCA ) = 180°

=> angle PAC + angle PCA = 90°

Now , in ∆ APC ,

angle PAC + angle PCA + angle APC = 180°

=> 90° + angle APC = 180°

=> angle APC = 90 °