Question

Please help me in this question ​

Answer 1

Hi mate here is the answer:--

Given:-

AB and CD are two parallel lines intersected by a transversal EF at G and H respectively.

GM is the bisector of ∠ AGH, GL is the bisector of ∠ BGH, HL is the bisector of ∠ DHG, HM is the bisector of ∠ CHG.

Proof:-

AB || CD and EF intersects them.

∠ AGH = ∠ GHD [Alternate angles]

1/2 ∠ AGH= 1/2 ∠ GHD

angle1=angle2.............(1)

But these angles form a pair of equal alternate angles for lines GM and HL and transversal GH.

⇒ GM || HL

Similarly HM || GL.

⇒ GLHM is a parallelogram.

⇒ AB || CD and EF is the transversal.

∴ ∠ BGH + ∠ GHD = 180° [ The sum of the interior angles on the same side of the transversal is 180° ]

⇒ 1/2∠ BGH + 1/2∠ GHD =180/2 = 90°

⇒ ∠ 3 + ∠ 2 = 90° ……………………….(ii)

In Δ GHL,

∠ 3 + ∠ 2 + ∠ GLH = 180° [Sum of the three angles of a triangle is 180° ]

⇒ 90° + ∠ GLH = 180°

⇒ ∠ GLH = 180° - 90° = 90°

⇒ One angle of the parallelogram is a right angle.

⇒ Parallelogram GLHM is a rectangle

Hope it helps you ❣️☑️☑️☑️