A geometry question involving a parallelogram within a triangle.

Question

The marked lengths are equal, what is the value of x?

Apologies if this problem is too easy for this website, yet I find this question to seriously challenge my mathematical capacity.

Answer 1

Let $\angle ABC=a$, then $\angle ACB=180-x-a$.

$BC\|DE$ so $\angle D=a,\angle E=180-x-a$.

Since $\triangle BDF,\triangle CEG$ are isosceles, $\angle DBF=a,\angle GCE=180-x-a$.

$\angle CGF=2(180-x-a)$ since it is an exterior angle of $\triangle CEG$.

Opposite angles of parallelogram are equal so $\angle FBC=\angle CGF$.

Now at $B,a+a+2(180-x-a)=180$, giving $x=90$.

Answer 2

Here is a diagram:

Let $\angle DEF = x$, so $\angle DEB = 180^{\circ} - x$. Then, $\angle GFE = 180^{\circ} - x$, which means $\angle GFA = x$. However, since $\triangle DBE$ is isosceles, we have that $\angle DEB = \frac{180 - \angle DEB}{2} = \frac{x}{2}$. Similarly, $\angle GAF = \frac{180 -\angle GFA}{2} = 90^{\circ} - \frac{x}{2}$. Thus, $\angle BCA = 180^{\circ} - \angle CBA - \angle CAB = 180^{\circ} - \angle DBE$$ - \angle GAF = \boxed{90^{\circ.}}$