In the figure shown, AB=AC and CE=CF. What is the value of x

Note that;
$$::\ Angle\ on\ a\ straight\ line\ =\ 180^o$$
$$::\ Sum\ of\ interior\ angles\ of\ a\ triangle=\ 180^o$$
::Angles opposite a congruent (the same) side are also congruent (the same)
:: Vertical angles are also the same (congruent)
Given that ;
$$Angle\ F\ =\ 40^o$$
$$Angle\ F\ =Angle\ E\ in\ \angle CEF\ \left(angles\ opposite\ congruent\ sides\right)$$
$$Therefore,\ Angle\ F\ =\ Angle\ E\ =40^o$$
In triangle CEF
$$Angle\ C\ +\ E\ +\ F=\ 180^o$$
$$Angle\ C\ +\ 40^o\ +\ 40^o=\ 180^o$$
$$Angle\ C\ +\ 80^o=\ 180^o$$
$$C\ =\ 180^o-80^o=100^o$$
The straight line FA is divided at C
$$Angle\ on\ a\ straight\ line\ FA=\ 180^o$$
$$Interior\ angle\ C\ in\ triangle\ \angle ACB=\ 100^o\ -\ 180^o=80^o\ $$
$$Angle\ C\ in\ triangle\ \ \angle ACB=80^o\ $$
Opposite side AB = AC (congruent sides and angles)
$$Angle\ B\ in\ \angle ACB\ -\ \angle\ ACB\ =\ 80^o$$
$$Angle\ E\ in\ \angle CEF\ =\ \angle BED\ =\ 40^o\left(Vertical\ angles\right)$$
In triangle BDE
$$Angle\ B\ +\ D\ +\ E\ =\ 180^o$$
$$Angle\ 80^o\ +\ D\ +\ 40^o=\ 180^o$$
The straight line BA is divided at D ,
$$Angle\ on\ a\ straight\ line\ BA\ =\ 180^o$$
$$x^o=\ 180^o-\ 60^o$$
$$x^o=\ 120^o$$
Option C is CORRECT.

fskilnik@GMATH wrote:In the figure shown, AB=AC and CE=CF. What is the value of x?

(A) 90
(B) 110
(C) 120
(D) 130
(E) 140



Source: https://www.GMATH.net

HI deloitte247,

Thank you for your solution!

The "alternate" way, presented in the figures below, has (I believe) one merit:

It takes into account the "exterior angle property", avoiding some intermediate calculations.




This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.