# World's Hardest Easy Geometry Problem

## Theorem

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 $CG$.

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:

- $\triangle ACG = \triangle CAE$

and so:

- $(1): \quad CE = AG$

Now we have that:

- $\angle AGB = \angle DGF = 60 \degrees$
- $DG = GF$

So $\triangle DGF$ 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

\(\ds AG\) | \(=\) | \(\ds CE\) | from $(1)$ | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds EF\) | \(=\) | \(\ds GF\) | as $AF = CF$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds EF\) | \(=\) | \(\ds DF\) | as $\triangle DFG$ is equilateral | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \angle FED\) | \(=\) | \(\ds \angle FDE\) | as $\triangle DFE$ is isosceles |

From Parallelism implies Equal Corresponding Angles:

- $\angle DFE = \angle ABC = 80 \degrees$

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:

- $\angle DEA = \angle FED - \angle AEB$

\(\ds \angle DEA\) | \(=\) | \(\ds \angle FED - \angle AEB\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 50 \degrees - 30 \degrees\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 20 \degrees\) |

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

Hence the result.

$\blacksquare$

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