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


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 \(\angle BAD\).


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, \(\angle BAD=100^\circ\).