Question

A transversal intersects two parallel lines. Prove that the bisectors of any pair of co interior angles intersects to form a right angled triangle

Answer 1

Step-by-step explanation:

Let line AB be parallel to CD, so, AB || CD

Now, let QR be the Transversal, and their intersections be at O and P on AB and CD respectively.

Now,

Co-interior angles are ∠AOP and ∠CPO,

∠BOP and ∠DPO

So, let's choose ∠BOP and ∠DPO

we know that, ∠BOP + ∠DPO = 180° (Co interior angles)

Let x = ∠BOP and y = ∠DPO

thus,

x + y = 180° ----- 1

so, their bisectors will be ∠POM and ∠MPO

So, ∠POM = x/2 and ∠MPO = y/2

Now in ΔOMP

∠POM + ∠MPO + ∠PMO = 180° (Angle Sum Property)

so, (x/2) + (y/2) + ∠PMO = 180°

(x + y)/2 + ∠PMO = 180°

Thus,

∠PMO = 180° - (x + y)/2

From eq.1 we get,

∠PMO = 180° - (180/2)°

∠PMO = 180° - 90° = 90°

∠PMO = 90°

Hence, ΔOMP is a right triangle

Thus, the bisectors of any pair of co interior angles intersects to form a right angled triangle when two parallel lines are intersected by a transversal.

Hope it helped and you understood it........All the best