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In a triangle $\text{ABC}, \angle \text{BCA} = 50^{\circ}. \text{D}$ and $\text{E}$ are points on $\text{AB}$ and $\text{AC},$ respectively, such that $\text{AD = DE}.$ If $\text{F}$ is a point on $\text{BC}$ such that $\text{BD = DF},$ then $\angle \text{FDE, in degrees},$ is equal to

  1. $96$
  2. $72$
  3. $80$
  4. $100$
in Quantitative Aptitude 2.7k points 5 79 342
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Let’s draw the diagram.

Let, the $\angle \text{DAE} = \angle \text{AED} = a^{\circ},$ and $\angle \text{DFB} = \angle \text{FBD} = b^{\circ},$ and $\angle \text{FDE}=x^{\circ}.$

Now,

  • $\angle \text{ADE} = 180^{\circ} – 2x^{\circ}$ 
  • $\angle \text{BDF} = 180^{\circ} – 2y^{\circ}$ 

We know that, angle on straight line equal to $180^{\circ}.$

So, $180–2a+x+180-2b=180$

$\Rightarrow x-2a-2b+180=0$

$\Rightarrow x=2a+2b-180 \; \longrightarrow (1) $

In $\triangle \text{ABC}, a+b+50=180 \quad [\because \text{Sum of the angles of triangle} = 180^{\circ}]$

$\Rightarrow a+b=130$

Put the value of $a+b$ in equation $(1).$

$\Rightarrow x=2(a+b)-180$

$\Rightarrow x=2(130)-180$

$\Rightarrow x=260-180$

$\Rightarrow \boxed{x=80^{\circ}}$

$\therefore$ The $\angle \text{FDE},$ in degree, is equal to $80^{\circ}.$  

Correct Answer $:\text{C}$

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