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In a triangle $ABC, AB=AC$, $BC=22\sqrt3$, $\cos A=-\frac{17}{225}$. $D$ is a point on $AC$ such that $AD<DC$ and $P$ is the point on the segment $BD$ such that $\angle APC = 90^{\circ}$. Given $\angle ABD=\angle BCP$, find $BD$.

Here's a diagram of what it's like I guess:

triangle_diagram

With reference to the diagram, $\cos (x+y)=\frac{11}{15}$, after taking $\cos (180^{\circ}-A)$ and so on. I kind of did cosine rule for both triangles $ABD$ and $BCD$ to get expressions for $BD^2$ and tried to sub in $\cos(x+y)$ but that is quite messy and doesn't really get anything. I also tried Stewart's theorem for $BD^2$. Are there any similar triangles or auxiliary lines that I'm not seeing? Can someone give a hint on this? Thanks.

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2 Answers 2

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Hint:

\begin{align} \triangle BCD&\cong \triangle CPD ,\\ \frac{|BD|}{|CD|} &= \frac{|CD|}{|PD|} =\frac{|BC|}{|PC|} \end{align}

\begin{align} \cos\tfrac\alpha2&=\frac{2\sqrt{26}}{15} =\sin\beta ,\\ \cos\beta&=\frac{11}{15} ,\\ |AB|=|AC|&=\frac{|BC|}{2\cos\beta} =15\sqrt3 ,\\ |AM|=|MC|=|MP|&=\tfrac12|AC|=\tfrac{15\sqrt3}2 ,\\ \triangle CDP:\quad \frac{|CD|}{|PD|} &=\frac{\sin\beta}{\sin(\beta-\phi)} ,\\ \triangle PMD:\quad |PD| &= \frac{|PC|\sin(\beta-\phi)}{\sin(2\beta-\phi)} ,\\ \frac{|BD|}{|CD|} &=\frac{|BC|}{|PC|} \\ &=\frac{|BC|}{|AC|\cos(\beta-\phi)} =\frac{22}{15\cos(\beta-\phi)} . \end{align}

From \begin{align} \frac{\sin\beta}{\sin(\beta-\phi)} &=\frac{22}{15\cos(\beta-\phi)} ,\\ \phi& =\beta-\arctan(\tfrac{15\sin\beta}{22}) =\beta-\arctan(\tfrac{\sqrt{26}}{11}) \end{align}

\begin{align} \frac{|BD|}{|CD|} &=\frac{22}{15\cos(\arctan(\tfrac{\sqrt{26}}{11}))} =\frac{14\sqrt3}{15} ,\\ \triangle CDP:\quad |PD| &= \frac{|AC|\cos(\beta-\phi)\sin(\beta-\phi)}{\sin(2\beta-\phi)} . \end{align}

After simplification, $|PD|=\tfrac{75}7$ and then

\begin{align} \frac{|BD|}{|CD|} \cdot \frac{|CD|}{|PD|} &=\left(\frac{|BD|}{|CD|}\right)^2 =\frac{196}{75} =\frac{|BD|}{|PD|} ,\\ |BD|&=\frac{196}{75} \cdot\frac{75}7 =28 . \end{align}

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  • $\begingroup$ Ok I see, thank you. $\endgroup$ Commented Jun 20, 2018 at 10:13
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We indicate with $\alpha$ the angle $\angle BAC$, so $cos(\alpha)=-\frac{17}{225}$. The angle $\angle ABC=\frac{\pi}{2}-\frac{\alpha}{2}$ and the angle $\angle PCA=\angle PBC= (\frac{\pi}{2}-\frac{\alpha}{2})-x$ but $APC$ is rectangle and so $\angle PAC=\frac{\pi}{2}- \angle PCA= \frac{\pi}{2}- (\frac{\pi}{2}-\frac{\alpha}{2})+x$ $= \frac{\alpha}{2}+x $

So

$\angle PAD= \frac{\alpha}{2}+x $ And $\angle PAB=\alpha-(\frac{\alpha}{2}+x)= \frac{\alpha}{2}-x$ And $\angle PDA=\angle BDA=\pi-\alpha-x$

$PD=\frac{sin(\frac{\alpha}{2}+x)}{sin(\pi-\alpha-x)}AP= \frac{sin(\frac{\alpha}{2}+x)}{sin(\alpha+x)}AP=$ $ \frac{sin(\frac{\alpha}{2}+x)}{sin(\alpha+x)}sin(\frac{\alpha}{2}+x)AC= \frac{sin^2(\frac{\alpha}{2}+x)}{sin(\alpha+x)}AC= $ $\frac{sin^2(\frac{\alpha}{2}+x)}{sin(\alpha+x)}sin(\frac{\pi}{2}-\frac{\alpha}{2})BC $

and

$BP=\frac{sin(\frac{\alpha}{2}-x)}{sin(x)}AP= \frac{sin(\frac{\alpha}{2}-x)}{sin(x)} sin(\frac{\alpha}{2}+x)AC= $ $\frac{sin(\frac{\alpha}{2}-x)}{sin(x)} sin(\frac{\alpha}{2}+x)sin(\frac{\pi}{2}-\frac{\alpha}{2})BC $

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