parmenides51 wrote: In triangle $ABC$ it is known that $\angle ACB = 2\angle ABC$. Furthermore $P$ is an interior point of the triangle $ABC$ such that $AP = AC$ and $PB = PC$. Prove that $\angle BAC = 3 \angle BAP$. Be $\angle ABC=2\theta$ and $\angle ACP=\omega$ $\frac{sin2\omega}{sin\omega}=\frac{PC}{AP}=\frac{BP}{AP}=\frac{sin(2\omega-3\theta)}{sin(\omega-\theta)}\Rightarrow 2cos\omega.sin(\omega-\theta)=sin(2\omega-3\theta)\Rightarrow sin(2\omega-\theta)-sin\theta=sin(2\omega-3\theta)$ $\Rightarrow 2cos(2\omega-2\theta)sin\theta=sin\theta \Rightarrow 2\omega-2\theta=60^{\circ} $

Hint

Best regards, sunken rock

Let angle ABC=x angle ACB =2x angle BAC=180°-3x AP=AC Let angle APC=angle ACP =y angle PAC=180°-2y angle BAP =2y-3x angle ABP=y-x[PB=PC] In ∆ APB, by sine rule, PB/ AP=sin(2y-3x)/ sin(y-x) ∆ APC, by sine rule, PC/ AP=sin 2y/ sin y=2cos y These two ratios are equal, sin(2y-3x)=2sin(y-x)cos y sin(2y-3x)=sin(2y-x)+sin x -2cos(2y-2x) sin x-sin x=0 sin x{ -2cos(2y-2x)-1}=0 sin x≠0 -2cos(2y-2x)-1=0 cos(2y-2x)=cos 60° 2y-2x=60°, y-x=30°, y=x+30° angle BAP =2y-3x=60°-x 3 (angle BAP)=180°-3x angle BAC=180°-3x So, angle BAC=3(angle BAP) PROVED # KRISHIJIVI

Take $A'$ reflection of $A$ about the perpendicular bisector of $BC$, notice that $AA'BC$ is an isosceles trapezoid and $A'B=AA'=AC$, so $A'$ is the circumcenter of $\triangle ABP$, that is, $m(\widehat{BA'P})=2m(\widehat{BAP})$, but by symmetry $\widehat{BA'P}=\widehat{CAP}$, done. Best regards, sunken rock

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