Problem 3

On the sides of an arbitrary triangle

ABC

, triangles

ABR

BCP

CAQ

are constructed externally with

\angle CBP = \angle CAQ = 45^\circ

\angle BCP = \angle ACQ = 30^\circ

\angle ABR = \angle BAR = 15^\circ

. Prove that

\angle QRP = 90^\circ

and

QR = RP

Step 5 of 6: Simplify with the triangle identities

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PR^2-QR^2=0,\qquad PR^2+QR^2-PQ^2=0

Detailed analysis

Substitute the sine-rule values from Step 2 into Step 4, expand $\cos(X+60^\circ)$ , and use the sine and cosine rules in $ABC$ : $a\sin B=b\sin A$ , $ac\cos B=(a^2+c^2-b^2)/2$ , and $bc\cos A=(b^2+c^2-a^2)/2$ . The terms cancel to give $PR^2=QR^2$ and $PR^2+QR^2=PQ^2$ .