euclidean-geometrygeometryplane-geometry
For reference: In the interior of a triangle ABC, a point $P$ is marked in such a way that: $PC=BC$ and the measure of the angle $PAB$ is equal to the measure of the angle $PAC$ which is $17°$. calculate the measure of angle $PCB$, if the measure of angle $B=107^o$ (Answer:$26^o$)
My progress
$\triangle ABC: \angle C = 180-107-34 = 124+\theta\\
\angle CBP=\angle CPB=90^o – \frac{\theta}{2}\\
\triangle APC: \angle APC = 124^o+\theta\\
\triangle ABP: \angle BPA = 146-\frac{\theta}{2} $
…?
Many thanks to @Ivan for identifying the error in the original image, and here is a solution by geometry as per the request of OP.
Extend $\small PH$ to meet $\small BC$ extend at $\small G$. $\angle CGH=53^\circ$.
$\small \therefore\angle BGP=\angle BAP\implies AGBP\ \text{is cyclic}$
As $\small AB=AP$, $\small \angle ABP=63.5^\circ$, and so is $\small \angle AGP$ because they are in the same segment.
Therefore, $\small \triangle AHG$ is a special right triangle with perpendicular sides in the ratio 1:2.
$\small \therefore HG=6$
Similarly, as $\small \triangle CGH$ is a $\small 3:4:5$ right triangle, $\small HC=8$.
$I$ and $J$ are incenters of $\triangle APB$ and $\triangle BPC$ respectively.
As $MI$ is parallel to $BP$ and $I$ is incenter,
$\displaystyle \frac{BM}{MX} = \frac{PI}{IX} = \frac{BP}{BX}$
and similarly, $ ~ \displaystyle \frac{BN}{NY} = \frac{PJ}{JY} = \frac{BP}{BY}$
So if we can show that $BX = BY$, that would show $BM = BN$ and we are done.
By some angle chasing, we can show that $\triangle PBX \sim \triangle PCY$ and I leave this as an exercise for you.
Then, $ \displaystyle \frac{BX}{BP} = \frac{CY}{CP}$
But given $PY$ is angle bisector of $\angle BPC$,
$ \displaystyle \frac{CY}{CP} = \frac{BY}{BP}$
$ \therefore BX = BY ~ $ and that completes the proof that $BM = BN$.
Best Answer
No trigonometric function is used in this answer.
As shown in the figure above, we start by constructing $\triangle APD$ such that $\triangle APD\cong\triangle APC$.
Since $\triangle ACD$ is an isosceles, from $\angle DAC=34^\circ$ we have $$\angle ADC=\angle ACD=73^\circ$$ It's given that $\angle ACB=39^\circ$, so $$\angle BCD=\angle ACD-\angle ACB=34^\circ$$
Here you might already know where this is going. With $\angle BCD=34^\circ$, we obtain $$\angle CBD=180^\circ-\angle BCD-\angle BDC=73^\circ$$ and it quickly follows that $$BC=CD$$ Now, with $BC=PC$ and $PD=PC$, we have an important conclusion, which is $$PC=CD=PD$$ Hence $\triangle PCD$ is an equilateral triangle, and $\angle PCD=60^\circ$. This implies $$\angle ACP=\angle ACD-\angle PCD=13^\circ$$ and finally, $$\angle PCB=\angle ACB-\angle ACP=\theta = 26^\circ$$ Hope this helps.