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
19.09.2018 07:17
We can easily prove $AC=BC=BP$, all with no trig! Best regards, sunken rock
19.09.2018 11:33
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 $a=b$ $Q.E.D.$
03.04.2020 10:20
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