In the figure, AB and CD are parallel lines. The bisectors of the interior angles on the same side of the transversal EF intersect at P. Prove that ∠GPH = 90°.
∠BGH + ∠DHG = 180° ….(Consecutive interior angles of AB || CD)….(i)
Now ray GP bisects ∠BGH
... ∠1 = ∠2 =
∠BGH ….(ii)
Similarly ∠3 = ∠4 = ∠DHG ….(iii)
Adding equations (ii) and (iii), we get
... ∠1 + ∠3 = ∠BGH + ∠DHG
... ∠1 +∠3 = (∠BGH + ∠DHG) ….(iv)
From equations (i) and (iv), we get
... ∠1 + ∠3 = x 180° = 90°
Now in ΔPGH, we have
... ∠1 + ∠3 + ∠GPH = 180° ….(sum of the angles of Δ)
⇒ 90° + ∠GPH = 180°
⇒ ∠GPH = 180° - 90° = 90°
Now ray GP bisects ∠BGH
... ∠1 = ∠2 =
∠BGH ….(ii)Similarly ∠3 = ∠4 =
∠DHG ….(iii)Adding equations (ii) and (iii), we get
... ∠1 + ∠3 =
∠BGH + ∠DHG... ∠1 +∠3 =
(∠BGH + ∠DHG) ….(iv) From equations (i) and (iv), we get
... ∠1 + ∠3 =
x 180° = 90°Now in ΔPGH, we have
... ∠1 + ∠3 + ∠GPH = 180° ….(sum of the angles of Δ)
⇒ 90° + ∠GPH = 180°
⇒ ∠GPH = 180° - 90° = 90°
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