Chapter 3, Problem 13CT

To determine

To prove: $\overleftrightarrow{BF}$ is parallel to $\overleftrightarrow{DG}$ in the given figure.

Explanation of Solution

Given information:

  

  $\overleftrightarrow{AB}\Vert \overleftrightarrow{CD}$

  $\overrightarrow{BF}$ bisects $\angle ABE$ , i.e. $\angle ABF=\angle FBE$ ------------------- (1)

  $\overrightarrow{DG}$ bisects $\angle CDB$ i.e. $\angle CDG=\angle GDB$ ------------------- (2)

Proof:

  

Here, $\angle CDX=\angle ABD$ ------------------- (3)

[Corresponding angles, since $\overleftrightarrow{AB}\Vert \overleftrightarrow{CD}$ ]

Now,

  $\angle CDX+\angle CDG+\angle GDB={180}^{\circ }$ ------------------- (4)

  $\angle ABD+\angle ABF+\angle FBE={180}^{\circ }$ ------------------- (5)

From equation (4) and (5)

  $\angle CDX+\angle CDG+\angle GDB=\angle ABD+\angle ABF+\angle FBE$

  $\Rightarrow \angle CDX+\angle CDG+\angle CDG=\angle ABD+\angle ABF+\angle ABF$

[Using (1) and (2)]

  $\Rightarrow 2\angle CDG=2\angle ABF$ [Using (3)]

  $\Rightarrow \angle CDG=\angle ABF$------------------ (6)

Again

  $\Rightarrow \angle CDG=\angle AYG$ ------------------- (7)[Corresponding angles, since $\overleftrightarrow{AB}\Vert \overleftrightarrow{CD}$ ]

From (6) and (7)

  $\Rightarrow \angle AYG=\angle ABF$ , which are corresponding angles

Hence, $\overleftrightarrow{BF}\Vert \overleftrightarrow{DG}$ as $\angle AYG=\angle ABF$ .