A geometry problem by Ossama Ismail

Geometry Level 2

In the square $ABCD$ , $E$ is the midpoint of $AB$ . A line is drawn from $E$ , perpendicular to $CE$ , and intersecting $AD$ at $F$ .

Find the ratio of $\angle BCF$ to $\angle BCE$ .

2 :

\frac{3}{2}

2 : 1 3 : 2 1 :

\frac{4}{5}

1 solution

Ron Gallagher
Apr 29, 2020

Without loss of generality, let the length of the side of the square be 2. Since E is the midpoint of AB, this means that AE = EB = 1. By the Pythagorean Theorem applied to triangle CEB, we find EC = Sqrt(5). Similarly, applying the Pythagorean Theorem to triangles AFE, FDC, and FEC, we find that AF = 1/2, FD = 3/2, and FE = (1/2) Sqrt(5). We then note that tan(angle BCE) = 1/2 and tan(angle ECF) = (1/2) sqrt(5) / sqrt(5) = 1/2. Therefore, angle BCE = angle ECF, so that angle BCF is twice the measure of angle BCE (i.e., the ratio is 2:1).

A geometry problem by Ossama Ismail

1 solution

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