Geometry Problem 1613: Square, Ratio 2:1, and the 53/2 Degree Angle Challenge

Diagram showing triangle ABC with perpendicular medians AM and CN intersecting at centroid G

Problem Statement:

In a square $ABCD$, let $E$ be a point on the extension of side $AB$, and $F$ be a point on the segment $CE$ such that $EF = 2CF$. A line passing through $F$ is perpendicular to $CE$ and intersects the extension of side $DA$ at point $G$. If $m\angle EGF = 53^\circ/2$, prove that: \[ m\angle BCE = 45^\circ \]
Technical Note:

In classical geometry, the angle \( \frac{53}{2}^\circ \) is the traditional shorthand for this configuration. Mathematically, the exact value is defined as:

\( \theta = \arctan\left(\frac{1}{2}\right) \approx 26.565^\circ \)

In a \( 3-4-5 \) triangle, this corresponds to the angle formed by the bisector of the larger acute angle and the adjacent leg, which naturally establishes the \( 1:2 \) ratio seen in this square's construction.

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Understanding the Problem

Analyzing the relationship between the 1:2 ratio ($EF=2CF$) and the notable angle of $53^\circ/2$ is key to finding the synthetic path to $45^\circ$.


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