Question 8
In the adjoining figure, $$ABCD$$ is a square and $$BE=AB$$. If the measure of $$\angle DFC$$ is $$x^\circ$$, then the value of $$2x$$ in degrees is
Solution
Since $$AB=BC=BE$$, triangle $$BEC$$ is isosceles. Also, $$\angle EBC=20^\circ+45^\circ=65^\circ$$, so $$\angle BCE=\angle BEC=\frac{180^\circ-65^\circ}{2}=57.5^\circ$$. Hence $$\angle FCD=90^\circ-57.5^\circ=32.5^\circ$$ and $$\angle FDC=45^\circ$$. Therefore, $$x=180^\circ-45^\circ-32.5^\circ=102.5^\circ$$, giving $$2x=205^\circ$$.
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