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Triangle $ABC$ is acute. Point $D$ lies on $\overline {AC}$ so that $\overline {BD}\perp \overline {AC}$, and point $E$ lies on $\overline {AB}$ such that $\overline {CE}\perp\overline {AB}$. The intersection of segments $\overline {CE}$ and $\overline {BD}$ is $P$. Find the value of $AE$ if $CP=10$, $PE=20$, and $EB=30$.

 Sep 17, 2019
 #1
avatar+104869 
+1

Here's a rough image  (not to scale)

 

 

Note that angle PEB  = angle CDP  = 90°

And angle EPB  = angle DPC

So triangles PEB  and   PDC  are similar

 

And  because BPE is a right triangle then  BP is the hypotenuse = sqrt (20^2 + 30^2) =  sqrt ( 1300)  = 10sqrt (13)

Then

BE / BP  = CD /CP                      and      PE /  BE  = PD / CD    

30/[10sqrt(13)]  = CD /10                        20 / 30  = PD / [30/sqrt (13)] 

30/sqrt(13)  = CD                                    20 / 30   =  sqrt(13)* PD /30

                                                                 PD = 20/sqrt(13)

 

Also angle PDC = angle CEA

And angle DCP = angle ACE

So  triangles  CEA  and CDP  are similar

 

So

 

AE / EC  = PD / DC

AE / 30  = [20/sqrt(13)] / [30/sqrt (13) ]

AE / 30   =  20 / 30

 

AE  = 20

 

 

cool cool cool

 Sep 18, 2019
 #2
avatar+23317 
+2

Triangle \(ABC\) is acute. Point \(D\) lies on \(\overline {AC}\) so that \(\overline {BD}\perp \overline {AC}\),
and point \(E\) lies on \(\overline {AB}\) such that \(\overline {CE}\perp\overline {AB}\).
The intersection of segments \(\overline {CE}\) and \(\overline {BD}\)is \(P\).
Find the value of \(\overline{AE}\)  if \(CP=10\), \(PE=20\), and \(EB=30\).

 

 

\(\begin{array}{|rcll|} \hline \tan(\alpha) = \dfrac{x}{20} &=& \dfrac{10+20}{30} \\\\ \dfrac{x}{20} &=& 1 \\\\ x &=& 20 \\\\ \mathbf{\overline{AE}} &=& \mathbf{20} \\ \hline \end{array}\)

 

laugh

 Sep 18, 2019

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