Olympiad question: In the regular pentagon $ABCDE$, the perpendicular at $C$ to $CD$ meets $AB$ at $F$. Prove that $AE + AF = BE$.

Extend the segment $CF$ to meet the line $AE$ at $G$. Easy angle chase we see that $AG = AF$ so we need to prove $EG = BE (= CE)$ and this is true since $\angle CEB = \angle BEG (= 36^{\circ})$ and $CG\bot BE$.

![enter image description here


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Note

$$\angle BAG = \angle ABG = 36,\>\>\>\>\> \angle EAG = \angle AGE = 72$$

and the triangles CBF and CGF are congruent, which leads to

$$\angle AFG = \angle AGF = 72$$

So, the triangles AEG, AGB and AFG are all isosceles, which yield

$$ BG = AG = AF,\>\>\>\>\>EG = EA$$

Thus,

$$BE = BG + GE = AF + AE$$

Tags:

Geometry

Euclidean Geometry

Geometric Construction

Plane Geometry

Contest Math

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