World's Hardest Easy Geometry Problem

Theorem

 * Worlds-hardest-easy-geometry-problem.png

Find $x$.

Solution
We are given that:


 * $\angle BAC = 70 \degrees + 10 \degrees = 80 \degrees$


 * $\angle ABC = 60 \degrees + 20 \degrees = 80 \degrees$

Thus by Triangle with Two Equal Angles is Isosceles, $\angle ABC$ is isosceles.

That is:
 * $AC = BC$

Using Sum of Angles of Triangle equals Two Right Angles we can construct some other angles:


 * $\angle ACB = 180 \degrees - 80 \degrees - 80 \degrees = 20 \degrees$


 * $\angle BDC = 180 \degrees - 20 \degrees - 20 \degrees = 140 \degrees$


 * $\angle AEC = 180 \degrees - 20 \degrees - 10 \degrees = 150 \degrees$


 * $\angle AEB = 180 \degrees - 70 \degrees - 80 \degrees = 30 \degrees$


 * $\angle ADB = 180 \degrees - 60 \degrees - 80 \degrees = 40 \degrees$

Construct $DF$ parallel to $AB$.

Draw $AF$ crossing $BD$ at $G$.

Draw $CG$.
 * Worlds-hardest-easy-geometry-problem-solution.png

Thus:
 * $AD = BF$
 * $\angle DAB = \angle FBA$
 * $BC$ is common

Hence from Triangle Side-Angle-Side Equality:
 * $\triangle ADB = \triangle AFB$

Immediately:
 * $\angle ABG = \angle BAG$

and so:
 * $\angle CAG = \angle CBG$

Thus:
 * $AC = BC$
 * $\angle CAG = \angle CBG$
 * $AG = BG$

Hence from Triangle Side-Angle-Side Equality:
 * $\triangle CAG = \triangle CBG$

Thus:
 * $\angle ACG = \angle BCG$

But as:
 * $\angle ACG + \angle BCG = 20 \degrees$

it follows that $\angle ACG = \angle BCG = 10 \degrees$

See that:
 * $\angle ACG = \angle CAE$
 * $\angle CAG = \angle ACE$
 * $AC$ is common

Hence from Triangle Angle-Side-Angle Equality:
 * $\triangle ACG = \triangle ACE$

and so:
 * $(1): \quad CE = AG$

Now we have that:
 * $\angle AGB = \angle DGF = 60 \degrees$
 * $DG = GF$

So $\triangle AGF$ is equilateral.

Hence:
 * $DF = DG = GF$

We have that:
 * $\angle ACF = 20 \degrees$
 * $\angle CAF = \angle CBD = 20 \degrees$

Hence from Triangle with Two Equal Angles is Isosceles:
 * $\triangle ACF$ is isosceles

So:
 * $AF = CF$

As $AG = CE$:
 * $EF = GF$

and so

We have that $\triangle DFE$ and $\triangle ABC$ are similar.

Thus:
 * $\angle DFE = \angle ABC = 80 \degrees$

Thus from Sum of Angles of Triangle equals Two Right Angles:
 * $\angle FED + \angle FDE = 180 \degrees - \angle DFE = 100 \degrees$

and so:
 * $\angle FED + \angle FDE = 50 \degrees$

Then:
 * $\angle DEA = \angle FED - \angle AEB$

But $\angle DEA$ is the angle $x$ which was to be found.

Hence the result.

Also see

 * Langley's Problem