Question

7. In the given figure, AB || CD. PA and PC are
bisectors of angle BAC and angleACD. Find angleAPC.
​

Answer 1

Answer:

Given A △ABC in which the bisectors of ∠B and ∠C meet the sides AC and AB at D and E respectively.

To prove AB=AC

Construction Join DE

Proof In △ABC, BD is the bisector of ∠B.

∴

BC

AB

=

DC

AD

...........(i)

In △ABC, CE is the bisector of ∠C.

∴

BC

AC

=

BE

AE

.......(ii)

Now, DE∣∣BC

⇒

BE

AE

=

DC

AD

[By Thale's Theorem]......(iii)

From (iii), we find the RHS of (i) and (ii) are equal. Therefore, their LHS are also equal i.e.

BC

AB

=

BC

AC

⇒ AB=AC

Hence, △ABC is isosceles.

Answer 2

Answer:

150°

Step-by-step explanation:

Alternate interior angle

∠BAC = ∠ACG = 120°

∠ACF + ∠FCG = 120°

So, ∠ACF = 120° – 90°

= 30°

Linear pair,

∠DCA + ∠ACG = 180°

∠x = 180° – 120°

= 60°

∠BAC + ∠BAE + ∠EAC = 360°

∠CAE = 360° – 120° – (60° + 30°)

= 150°