In triangle $ABC$ given is point $P$ such that $\angle ACP = \angle ABP = 10^{\circ}$, $\angle CAP = 20^{\circ}$ and $\angle BAP = 30^{\circ}$. Prove that $AC=BC$
Problem
Source: Regional Olympiad - Federation of Bosnia and Herzegovina 2018
Tags: geometry, Triangle, isosceles
sunken rock
19.09.2018 07:17
We can easily prove $AC=BC=BP$, all with no trig! Best regards, sunken rock
Tintarn
19.09.2018 11:33
Clearly everything is uniquely determined, so it suffices to work with the converse: Let $ABC$ be a triangle with angles $50^\circ, 50^\circ, 80^\circ$, isosceles at $C$. Let $P$ be the unique point inside the triangle such that $\angle BAP=30^\circ, \angle PBA=10^\circ$. Then we only need to prove that $\angle $ACP=10^\circ$.
Let $D$ be the point of reflection of $C$ over the line $AB$. Then $D$ is the circumcenter of triangle $APB$. So $\angle BDP=2\angle BAP=60^\circ$ whence triangle $BDP$ is equilateral.
Let $E \ne B$ be the point on $AB$ such that $EPB$ is isosceles at $P$. Then $\angle $EPA=20^\circ$ and $EP=PB=BD=AC$, so triangles $APE$ ad $PAC$ are congruent. But then $\angle ACP=\angle AEP=10^\circ$. Done.
dikhendzab
01.04.2020 21:17
Here is quick and nice trig solution. From sine law in triangle $ACP$ we get: $\frac{b}{\sin150^\circ}=\frac{AP}{\sin10^\circ}$ or $b=\frac{AP\sin150^\circ}{\sin10^\circ}=\frac{AP}{2\sin10^\circ}$ From sine law in triangle $ABP$ we get: $\frac{c}{\sin140^\circ}=\frac{AP}{\sin10^\circ}$ or $c=\frac{AP\sin140^\circ}{\sin10^\circ}$ Now, from cosine law and using that $\cos50^\circ=\sin40^\circ=\sin140^\circ$, $a^2=b^2+c^2-2bc\cdot \cos \angle BAC$ we will have $a^2-b^2=c^2-2bc\cdot \cos50^\circ=\frac{AP^2\sin^2 140^\circ}{\sin^2 10^\circ}-\frac{2AP^2 \sin140^\circ \cos50^\circ}{2\sin^2 10^\circ}=\frac{AP^2 \sin^2 140^\circ}{\sin^2 10^\circ}-\frac{AP^2 \sin^2 140^\circ}{\sin^2 10^\circ}=0$ $\implies$ $a^2-b^2=0$, $a^2=b^2$ and finally $a=b$ $Q.E.D.$
Andrew.xyz
03.04.2020 10:20
$\angle APC + \angle BAP = 180^{\circ}$ .
$\angle ABP + \angle PAC = \angle BAP$.
Let $K$ be the point inside $\bigtriangleup ABC$ such that $\angle PBK = 20^{\circ}, \angle BPK = 10^{\circ}$. So we got $\angle BKP = 150^{\circ}$ and $\angle ABK = \angle BAP = 30^{\circ}$. Hence we got that $ABKP$ is equilateral trapezoid, so $BK = AP$ and triangle $APC$ is congruent to triangle $BPK$, so $PK = PC$. By simple angle chasing we got $\angle KPC = 60^{\circ}$, so triangle $KPC$ is equilateral, so $KC = PC$. But now we easily can get that $\angle BKC = 150^{\circ}$, moreover, $BK = AP$ and $KC = PC$. So $\bigtriangleup BKC$ is congruent to $\bigtriangleup APC$, $AC = BC$. Done.
Krishijivi
13.07.2023 17:28
In∆APB, by sine rule sin 30°/BP= sin 10°/AP in ∆ APC , by sine rule, sin 10°/AP=sin 150°/AC sin 10°/AP=sin 30°/AC So, AC=BP Let angle PBC=x° So, angle PCB=110°-x In ∆BPC , by sine rule sin x/PC=sin(110°-x)/BP ∆ APC , by sine rule, sin 20°/CP=sin 150°/AC AC/PC=1/2sin 20° So, sin(110°-x)/sin x=1/2sin20° Which gives, sin(40°-x)=0=sin 0° So, x=40° So, angle ABC=50°, angle BAC=50°, SO, AC=BC # KRISHIJIVI