Question

On the sides of an arbitrary triangle ABC, triangles BPC, CQA, and ARB are externally erected such that
$\angle PBC=\angle CAQ={45}^{\circ }$,
$\angle BCP=\angle QCA={30}^{\circ }$,
$\angle ABR=\angle BAR={15}^{\circ }$;

• A. $\angle QRP={90}^{\circ }$
• B. $QR=RP$
• C. $QR>RP$
• D. $\angle QRP>{90}^{\circ }$

Answer 1

Answer: (A), (B)

The correct options are
A $\angle QRP={90}^{\circ }$
B $QR=RP$

Denote by K and L the feet of perpendicular from the points P and Q to the lines BC and AC respectively.
Let M and N be the points on AB (ordered A-N-M-B) such that the triangle RMN is isosceles with angle $R={90}^o$
By sine theorem we have $\frac{BR}{BA}=\frac{sin{15}^o}{sin{45}^o}$. Since $\frac{BK}{BC}=\frac{sin{45}^{o}sin{30}^o}{cos{15}^o}$, we deduce that $MK\parallel AC$ and $MK=AL$. Similarly, $NL\parallel BC$ and $NL=BK$. It follows that the vectros $\overrightarrow{RN},\overrightarrow{NL}$, and $\overrightarrow{LQ}$ are the images of $vecRM,\overrightarrow{KP}$, and $\overrightarrow{MK}$ respectively under a rotation of ${90}^o$, and consequently the same holds for their sums $\overrightarrow{RQ}$ and $\overrightarrow{RP}$. Therefore, $QR=RP$ and angle $QRP={90}^o$.