Challenging Problems in Geometry: Problem 4-36

Problem 4-36

The following is a problem that appears in Challenging Problems in Geometry by Alfred S. Posamentier and Charles T. Salkind:

A line, $\overline{PQ}$, parallel to base $\overline{BC}$ of $\triangle ABC$, cuts $\overline{AB}$ and $\overline{AC}$ at $P$ and $Q$, respectively. The circle passing through $P$ and tangent to $\overline{AC}$ at $Q$ cuts $\overline{AB}$ again at $R$. Prove that the points $R$, $Q$, $C$, and $B$ lie on a circle.

Solution

The “book” solution is pretty fast and involves showing angle $\angle QRB$ is supplementary to angle $\angle C$. However, Curzon came up with a slightly longer alternate proof today that involved showing the other pair of opposite angles are supplementary. I thought it was creative enough for a 9 year old that I’d write it here for posterity. And yes, I do realize that it’s essentially re-proving the Tangent-Chord Angle Theorem using facts about cyclic quadrilaterals :)

Curzon Solution

The key idea is to reflect point $Q$ across the circle to form diameter $\overline{QS}$. Then connect $\overline{QR}$ and $\overline{QS}$:

Now call angle $\angle B = \beta$ and do some basic angle chasing:

• $\angle APQ = \beta$ because $\overline{AB}$ is a transversal through parallel lines $\overline{PQ}$ and $\overline{BC}$
• $\angle QPR = 180° - \beta$ because they’re supplementary
• $\angle RSQ = \beta$ because $QPRS$ is cyclic and opposite angles are supplementary
• $\angle QRS = 90°$ because of inscribed angle theorem
• $\angle RQS = 90° - \beta$ because angles in triangle $RQS$ add to $180°$
• $\angle SQC = 90°$ because a radius (or diameter) makes a right angle with tangent lines

Lastly, $\angle RQC = \angle RQS + \angle SQC = (90° - \beta) + 90° = 180° - \beta$, which is the supplement of $\angle RBC = \beta$. $\square$