Question

In the given figure, line $DEparallel$line $GF$ ray $EG$ and ray $FG$ are bisectors of $\mathrm{\angle }DEF$ and $\mathrm{\angle }DFM$ respectively. Prove that,
$\left(1\right)$$angleDEG={\displaystyle \frac{1}{2}}\mathrm{\angle }EDF$
$\left(2\right)$$EF=FG$

Answer 1

Hint: In this question we will use the following properties:
If a line passes through the center of an angle then it divides the same angle into equal parts.
Sum of two interior angles of a triangle is equal to the opposite exterior angle of a triangle. Sum of all angles of the triangle is always equal to ${180}^{\circ }$. Two triangles are said to be similar if the corresponding angles of two triangles are congruent and lengths of corresponding sides are proportional. Two triangles are said to be congruent if all the sides of one triangle are equal to the corresponding sides of another triangle and the corresponding angles are equal.

Complete step-by-step answer:
Given, $DE\parallel GF$
Ray $EG$and ray $FG$ are bisectors of $\mathrm{\angle }DEF$and$\mathrm{\angle }DFM$.
Means $EG$and $FG$divides $\mathrm{\angle }DEF$and$\mathrm{\angle }DFM$respectively in two equal parts. So,
$\Rightarrow \mathrm{\angle }DEG=\mathrm{\angle }GEF={\displaystyle \frac{1}{2}}\mathrm{\angle }DEF$ ……..$1$
Also, $\mathrm{\angle }DFG=\mathrm{\angle }GFM={\displaystyle \frac{1}{2}}\mathrm{\angle }DFM$……. $2$
Since, $DE\parallel GF$
So, $\mathrm{\angle }EDF=\mathrm{\angle }DFG$ ………$3$ ($\because$we know that alternate interior angles are equal to each other)
Also we can write, $\mathrm{\angle }EDF={\displaystyle \frac{1}{2}}\mathrm{\angle }DFM$$.....4$
Now, consider $△DEF$
We know that the exterior angle of a triangle is equal to the sum of opposite two interior angles.
Therefore,
$\mathrm{\angle }DFM=\mathrm{\angle }DEF+\mathrm{\angle }EDF$
From equation $4$$\mathrm{\angle }EDF={\displaystyle \frac{1}{2}}\mathrm{\angle }DFM$. We can also write this equation like this, $2\mathrm{\angle }EDF=\mathrm{\angle }DFM$ Therefore,
$\Rightarrow 2\mathrm{\angle }EDF=\mathrm{\angle }DEF+\mathrm{\angle }EDF$
We get, $\mathrm{\angle }EDF=\mathrm{\angle }DEF$
From equation $1$, $\mathrm{\angle }DEG={\displaystyle \frac{1}{2}}\mathrm{\angle }DEF$. We can also write this equation like this $2\mathrm{\angle }DEG=\mathrm{\angle }DEF$
$\Rightarrow \mathrm{\angle }EDF=2\mathrm{\angle }DEG$
So we get $\mathrm{\angle }DEG={\displaystyle \frac{1}{2}}\mathrm{\angle }EDF$.
Hence proved.
Given, $DEparallelGF$
$\mathrm{\angle }DEG=\mathrm{\angle }EGF$…… $5$ ($\because$We know that alternate interior angles are equal to each other)
Now, From equation 1 , $\mathrm{\angle }DEG=\mathrm{\angle }GEF$
$\mathrm{\angle }GEF=\mathrm{\angle }EGF$
Since, the $△EGF$sides opposite to equal angles are also equal.
$\Rightarrow EF=FG$
Hence, proved

Note:While solving this question one should have remembered all the properties angles and triangles i.e. If a triangle has the same three angles then it has similar length of sides or vice versa and Bisectors always cut angle into two equal parts etc. Also should take care while doing calculation.