Find the $\angle ACB$ of $\triangle ABC$.

angleeuclidean-geometrygeometryreflectiontriangles

If $PC=2BP$, $\angle ABC= 45^\circ$, and $\angle APC=60^\circ$, find $\angle ACB$.

All solutions are acceptable but please try solving using reflection of point $C$ through the line segment $AP$.

I found a solution by adding a dot named D but in the given solution which I've drawn, I didn't understand how alpha was found. Although it is the original given solution in the geometry book.

enter image description here

It was only pointed that
$\angle ACB = \angle ADP = \frac{1}{2}(180^\circ – \angle BDP) = 75^\circ$.
Please explain this for me because this solution is blurry and vague for me.

Best Answer

Sometimes, geometry consists of dropping the right line or introducing the accurate point...

enter image description here Denote by $D$ the point on $AP$ such that $\angle PDC=90°$ and let $BP=x$. Note now that the triangle $\Delta PCD$ is a $30°-60°-90°$ triangle, which implies that $PD=x$.

Therefore $\Delta BPD$ is isosceles and hence $\angle PBD=30° \Rightarrow \angle DBA=15°$.

Easy angle chasing leads to the conclusion that $DC=DB=DA$, which implies that $\Delta CDA$ is also isosceles. Thus $\angle ACD=45°$. $$\therefore \angle ACB=45°+30°=75° \blacksquare$$

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