Question

AB and CD are two parallel lines intersected by a transversal EF.Bisectors of interior angles BMN and DNM on the same side of transversal intersect at P.Prove that angleMPN=90 degrees

Answer 1

Solution:
∠ BMN = ∠ DNM = 180° (Consecutive interior angles of AB II CD) ... (1)
Now ray MP bisects ∠ BMN
∴ ∠ 1 = ∠ 2 = 1/2 of ∠ BMN ...(2)
Similarly ∠ 3 = ∠ 4 = 1/2 of ∠ DNM ...(3)
Adding equations (2) and (3), we get
∴ ∠ 1 + ∠ 3 = 1/2 ∠ BMN + 1/2 ∠ DNM
∴ ∠ 1 + ∠ 3 =1/2(∠ BMN + ∠ DNM)... (4)
From equations (1) and (4), we get
 ∴ ∠ 1 + ∠ 3 = 1/2 × 180° = 90°
Now in Δ PMN, we have
∴ ∠ 1 + ∠ 3  ∠ MPN = 180° ..... (Sum of the angles of Δ)
⇒ 90° + ∠ MPN = 180°
⇒ ∠ MPN = 180 - 90°
⇒ ∠ MPN = 90° Hence proved.

Answer 2

Answer:

Step-by-step explanation:

Solution:

∠ BMN = ∠ DNM = 180° (Consecutive interior angles of AB II CD) ... (1)

Now ray MP bisects ∠ BMN

∴ ∠ 1 = ∠ 2 = 1/2 of ∠ BMN ...(2)

Similarly ∠ 3 = ∠ 4 = 1/2 of ∠ DNM ...(3)

Adding equations (2) and (3), we get

∴ ∠ 1 + ∠ 3 = 1/2 ∠ BMN + 1/2 ∠ DNM

∴ ∠ 1 + ∠ 3 =1/2(∠ BMN + ∠ DNM)... (4)

From equations (1) and (4), we get

 ∴ ∠ 1 + ∠ 3 = 1/2 × 180° = 90°

Now in Δ PMN, we have

∴ ∠ 1 + ∠ 3  ∠ MPN = 180° ..... (Sum of the angles of Δ)

⇒ 90° + ∠ MPN = 180°

⇒ ∠ MPN = 180 - 90°

⇒ ∠ MPN = 90° Hence proved.

Hope it will help you

THANKS

DR.EVA