# Find angle inside isosceles triangle

Please help. I have tried to find the answer but I couldn't. I have tried to draw a picture and measure the angle. It gives the answer $\theta = 25^\circ$ but I don't know how to do it.

Thank you

Question is not incomplete.

If we partition $\angle PBC$ into $5^\circ$ and $30^\circ$ as in the figure, $PB$ becomes angle bisector of $\Delta ABR$. Now, notice that $\angle BRP = 60^\circ$. Also notice that since $\Delta BRC$ is isosceles as well, line segment $AR$ divides $\angle BAC$ equally. So $\angle RAC = 50^\circ$ and from there, $\angle ARP = 60^\circ$. Now notice that $PR$ is also angle bisector of $\Delta ABR$. Since two angle bisectors of $\Delta ABR$ intersects at $P$, third must pass through $P$ as well. Therefore $AP$ is the third angle bisector and $\angle PAR = \theta$. So we have $2\theta = 50 \implies \theta = 25^\circ$.

EDIT: Proof for the statement "If two angle bisectors intersects at point $P$, then third angle bisector also passes through $P$":

Assume we know that $BP$ and $CP$ are angle bisectors of $\Delta ABC$. Then let us draw perpendicular line segments $PD$, $PE$ and $PF$. Now notice that $\Delta BFP \cong \Delta BDP$ since they have the same angles and they share the hypotenus $BP$ (this means their hypotenus are the same). Therefore, we have $|PD| = |PF|$. By similar argument, $\Delta CDP \cong \Delta CEP$ so $|PD| = |PE|$. So we have $|PD| = |PE| = |PF|$. Now, since $|PE| = |PF|$ and $\Delta AFP$ and $\Delta AEP$ shares hypotenus, we have $|AF| = |AE|$ by Pythagoras Theorem. So, we also have $\Delta AFP \cong \Delta AEP$, which implies $\angle FAP = \angle PAE$.

I'm glad that you asked for a proof because I thought proof could come from inscribed circle but now I noticed that we can't know whether $P$ is the center of inscribed circle or not just by knowing two angle bisectors insersect at $P$, directly (it can be derived as in the following but not obvious in the beginning). When we use the argument above, we can conclude that $P$ is the center of inscribed circle by using the equation $|PD| = |PE| = |PF| = r$ where $r$ is the radius of inscribed circle. So my hint when you first asked for a proof was result of this argument but wasn't quite right, I admit that.

• Bravo really nice solution. But I didn't say it's incomplete I said I can't find third relation I hope you understand – King Tut Jan 31 '18 at 12:28
• I wasn't referring you or @Gaurang Tandon, I just wanted to specify it. I'm sorry if I'm misunderstood. In any case, there are lots of olimpiad level geometry queations that would be incomplete if you try to solve them just by system of equations. Approaching a geometry question in that way is usually misleading. – ArsenBerk Jan 31 '18 at 12:32
• This is amazing! Thank you! :D – Gaurang Tandon Jan 31 '18 at 13:10
• Thank you very much. Very elegant solution but how can you know that if two angle bisectors intersects, the third line from the vortex to the intersection will be angle bisector too? Any for the proof of that? – Supphanat Jan 31 '18 at 15:57
• I edited my answer. Now it includes the proof you wanted. You're welcome and I thank you too for asking for a proof, I also learned it while trying :) – ArsenBerk Jan 31 '18 at 17:28