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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Also see

• Langley's Problem