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, 2015

Geometry Difficulty 5.8 AIME, harder Prove it Argentina

Points $D$ and $E$ divide side $AB$ of equilateral triangle $ABC$ into three equal parts; $D$ is between $A$ and $E$ . Point $F$ on side $BC$ is such that $CF = AD$ . Find the sum of the angles
$C\hat{D}F + C\hat{E}F.$

Solution

The conditions give $BF = BD$ ( $= \frac{2}{3}AB$ ), also $D\hat{B}F = 60^\circ$ , hence triangle $DBF$ is equilateral. Then
$DF \parallel AC \text{ as } B\hat{D}F = B\hat{A}C = 60^\circ.$
Hence $C\hat{D}F = A\hat{C}D$ .

On the other hand $A\hat{C}D = B\hat{C}E$ by the symmetry of the figure (or because triangles $ADC$ and $BEC$ are congruent). Then $C\hat{D}F = B\hat{C}E$ . So

the required sum $C\hat{D}F + C\hat{E}F$ is equal to $F\hat{C}E + C\hat{E}F$ . By the exterior angle theorem that last sum equals $B\hat{F}E$ . Now $FE$ is a median in the equilateral triangle $DBF$ , hence also a bisector. Therefore $B\hat{F}E = \frac{1}{2}B\hat{F}D = \frac{1}{2} \cdot 60^\circ = 30^\circ$ , which is the answer to the problem.

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