Question

in figure de is parallel to qr and ap and bp are bisectors of angle eab and angle rba respectively find angle apb

Answer 1

Given, angle DE parallel to QR and AB is transversal
So, angle EAB +angle ABE= 180 ( co -interior angle)
1/2 angle EAB+ 1/2 angle ABE= 180÷2 (multiplying 1/2 to both side)
So, angle ABP +angle BAP =90
Now, In triangle ABP,
(angle ABP+angle BAP)+angle APB = 180
90+angle APB= 180
angle APB = 180-90
=90

Answer 2

Answer:

Step-by-step explanation:

Given, angle DE parallel to QR and AB is transversal

So, angle EAB +angle ABE= 180 ( co -interior angle)

1/2 angle EAB+ 1/2 angle ABE= 180÷2 (multiplying 1/2 to both side)

So, angle ABP +angle BAP =90

Now, In triangle ABP,

(angle ABP+angle BAP)+angle APB = 180

90+angle APB= 180

angle APB = 180-90

=90