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

From the Iranian Geometry Olympiad, 2017:

In the regular pentagon $$ABCDE$$, the perpendicular at $$C$$ to $$CD$$ meets $$AB$$ at $$F$$. Prove that $$AE + AF = BE$$. Construction: https://www.geogebra.org/calculator/bnmgctmk

I can't seem to make much headway on this problem. You could probably use trigonometry to find the length of $$BE$$, but I am guessing there is a much easier (and elegant) solution that is eluding me.

• Have you seen this on the Art of Problem Solving? Jun 16, 2020 at 10:44

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 $$CG\bot BE$$.

• Which tool did you use for drawing this? Jun 18, 2020 at 9:18
• Geogebra............ Jun 18, 2020 at 10:58

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,

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

• Which tool do you use to draw figures like this? Jun 18, 2020 at 9:18

The answer described below does not use numerical values of angles. A couple of auxiliary lines is needed to be drawn in order to facilitate the proof. One of them is the line joining the two vertices $$A$$ and $$C$$, which intersects $$BE$$ at $$G$$. The other is $$FG$$.

Let the length of a side of the pentagon be $$a$$. Using the properties of a regular pentagon, we can state that $$BE$$ is parallel to $$CD$$, while $$AC$$ is parallel to $$DE$$. Thos makes $$CDEG$$ a parallelogram. However, because $$CD=DE=a$$ (two side of the pentagon) , $$CDEG$$ is an oblique equilateral parallelogram called a rhombus. Therefore, we have, $$EG=GC= a$$. Since $$AE$$ is also a sides of the pentagon, we can state, $$EG=AE \tag{1}$$

Furthermore, $$CB=CG$$, which confirms that $$BCG$$ is an isosceles triangle. Since $$CD$$ is parallel to $$BE$$, $$CF$$ is the perpendicular bisector of $$BG$$. Therefore, $$BGF$$ is also an isosceles triangle. Due to the prevalent symmetry of a regular pentagon, $$AG=BG$$, which makes $$ABG$$ is an isosceles triangle as well.

Let $$\measuredangle GAB=\phi$$ and $$\measuredangle FGA=\psi$$. Since $$ABG$$ is am isosceles triangle, we have $$\measuredangle ABG=\phi$$. Since $$BGF$$ is an isosceles triangle, $$\measuredangle BGF=\phi$$ as well. Consequently, $$\measuredangle AFG$$, which is one of the exterior angle the triangle $$BGF$$, is equal to $$2\phi$$. Now, by considering the sum of the three angles of the triangle $$AFG$$, we can write, $$\measuredangle GAF + \measuredangle AFG + \measuredangle FGA = 180^o \quad\rightarrow\quad 3\phi+\psi=180^o \tag{2}$$

Since $$AB$$ and $$EA$$ are two adjacent sides of the pentagon, $$ABE$$ is an isosceles triangle. Therefore, $$\measuredangle BEA=\measuredangle ABE=\phi$$. According to equation (1), $$AGE$$ is an isosceles triangle, which means $$\measuredangle AGE = 90^o-\frac{\phi}{2}$$. Now, we know three angles, which sum up to $$180^o$$, i.e., $$\measuredangle BGF + \measuredangle FGA + \measuredangle AGE=\phi+\psi+90^o-\frac{\phi}{2}=180^o \quad\rightarrow\quad \phi+2\psi=180^o \tag{3}$$

By subtracting equations (2) from (3), we can obtain the following relationship between $$2\phi$$ and $$\psi$$. $$\psi=2\phi$$

Therefore, $$AFG$$ is an isosceles triangle, where $$AF=AG$$. Bu we are already aware that $$AG=BG$$, because $$BGA$$ is an isosceles triangle. This means, $$BG=AF \tag{4}.$$

Now, we can prove the required relation using the equations (1) and (4) as shown below. $$EG+BG=AE+AF \quad\rightarrow\quad BE= AE+AF$$

• Which tool did you use in the construction of this figure? Jun 18, 2020 at 9:19
• I drew the figure in Microsoft Visio
– YNK
Jun 19, 2020 at 9:29

Algebraic solution: Let the side length of the pentagon be 1. Let $$x=\frac{\pi}5$$. Then the internal angles of the pentagon are $$3x$$ and we have $$\measuredangle CBE=2x$$, $$\measuredangle EBA=x$$

Then: $$BE=1+2\cos 2x,\quad AE+AF=2-\frac{\cos 2x}{\cos x}.$$ Hence: \begin{align} (BE-AE-AF)\cos x&=2\cos2x\,\cos x+\cos2x-\cos x\\ &=(\cos x+\cos3x)+\cos2x-\cos x\\ &=\cos3x+\cos2x=0. \end{align}

Let the side of the pentagon be $$1$$ and the point of intersection of $$CF$$ and $$BE$$ be $$P$$. We have $$\hat{EPC}=90$$ and $$\hat{ABE}=\hat{BEC}=36$$.

Then $$BE=2\cos36 = x = EC$$ $$\implies EP = x\cos 36\implies BP=x(1-\cos36)$$

Now, $$\frac{BP}{BF}=\cos36\implies BF=\frac{x(1-\cos36)}{\cos36}$$ $$\implies AF=1-\frac{x(1-\cos36)}{\cos36}$$ $$\implies AE+AF=2-\frac{x(1-\cos36)}{\cos36}=\frac{2\cos36-x+x\cos36}{\cos36}$$

But since $$2\cos36 = x$$, we have $$AE+AF=x=BE$$ QED

The diagonals of a regular polygon divide its vertex angles into equal parts. So we fold along line $$\overline{EB}$$:

From the angle equality noted above, the fold-projected point $$A'$$ lies on line $$\overline{EC}$$, and $$\overline{A'G}$$ which is perpendicular to $$\overline{CF'}$$ also bisects $$\angle CA'F'$$. That forces $$A'F'=A'C$$. Then

$$EC=EA'+A'C=EA'+A'F'=EA+AF$$

and from the regularity of the pentagon we also have $$EC=EB$$.