 # Problem of the Week Problem D and Solution The Angle Isn't Bad

## Problem

Ewan drew rhombus $$ABCD$$. Recall that a rhombus is a quadrilateral with parallel opposite sides, and all four sides of equal length. In Ewan’s rhombus, $$H$$ is on $$BC$$ in between $$B$$ and $$C$$, and $$K$$ is on $$CD$$ in between $$C$$ and $$D$$, such that $$AB=AH=HK=KA$$.

Determine the measure, in degrees, of $$\angle BAD$$. ## Solution

Since $$ABCD$$ is a rhombus, we know $$AB=BC=CD=DA$$. We’re also given that $$AB=AH=HK=KA$$. Let $$\angle ADK=x^\circ$$. Since $$AH=HK=KA$$, $$\triangle AHK$$ is an equilateral triangle and each angle in $$\triangle AHK$$ is $$60^\circ$$. In particular, $$\angle HAK =60^\circ$$.

In $$\triangle ADK$$, $$AD=AK$$ and so $$\triangle ADK$$ is isosceles. Therefore, $$\angle AKD=\angle ADK=x^\circ$$. Then $$\angle DAK=(180-2x)^\circ$$.

Since $$ABCD$$ is a rhombus, $$AB \parallel CD$$ and $$\angle ADC +\angle BCD=180^\circ$$. It follows that $$\angle BCD=(180-x)^\circ$$. But in the rhombus we also have $$BC \parallel AD$$ and $$\angle BCD +\angle ABC=180^\circ$$. It follows that $$\angle ABC=180^\circ-(180-x)^\circ=x^\circ$$.

In $$\triangle AHB$$, $$AH=AB$$ and so $$\triangle AHB$$ is isosceles. Therefore, $$\angle AHB=\angle ABH=x^\circ$$. Then $$\angle BAH=(180-2x)^\circ$$. Since $$ABCD$$ is a rhombus, $$BC\parallel AD$$, so \begin{aligned} \angle BAD &=180^\circ-\angle ABC\\ (180-2x)^\circ+60^\circ+(180-2x)^\circ&=180^\circ-x^\circ\\ (420-4x)^\circ&=(180-x)^\circ\\ 240^\circ&=(3x)^\circ\\ x^\circ&=80^\circ\end{aligned}

It follows that \begin{aligned} \angle BAD &=(180-x)^\circ\\ &=180^\circ-80^\circ\\ &=100^\circ\end{aligned}

Therefore, $$\angle BAD=100^\circ$$.