Question

If two parallel lines DE and QR are intersected by a transversal n, the bisectors of the two pairs of interior angles AP and BP intersect at P, prove that ∠APB = 90°.

Correct Answer= BRAINLIEST

Answer 1

eab plus rbe =180. angle on same sid of transversal
2x 1 + 2x 2 =180. bisector
so, 1+2=90
1+2+apb=180
apb =180-90
apb=90

hence proved

Answer 2

Given : DE and QR are parallel lines interested by transversal n . Intersecting CE at A and QR at B .AP and BP are bisectors  of angle EAB and angle RBA respectively

To Find :  prove that ∠APB = 90°.

Solution:

DE and QR are parallel lines interested by transversal AB

Properties of angles formed by transversal line  with two parallel lines :

• Corresponding angles are congruent.  

• Alternate angles are congruent.  ( Interiors & Exterior  both )  

• Interior angles are supplementary. ( adds up to 180°)

=> ∠EAB + ∠RBA = 180°

AP and BP are bisectors  of  ∠EAB and  ∠RBA

=> ∠BAP = (1/2)  ∠EAB

   ∠ABP = (1/2)  ∠RBA

=>  ∠BAP +   ∠ABP = (1/2)(∠EAB + ∠RBA )

=> ∠BAP +   ∠ABP = (1/2)(180° )

=>  ∠BAP +   ∠ABP =  90°

∠BAP +   ∠ABP =  + ∠APB = 180°   Sum of angles of triangle

=>  90°  + ∠APB = 180°

=> ∠APB = 90°

QED

Hence proved

Learn More:

7. If a transversal intersects two lines parallel lines then prove that ...

brainly.in/question/18440350

If two interior angles on the same side of a transversal intersecting ...

brainly.in/question/15906419