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 $EB=30$.
+128407 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
+26367 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 \(EB=30\).
\(\begin{array}{|rcll|} \hline \tan(\alpha) = \dfrac{x}{20} &=& \dfrac{10+20}{30} \\\\ \dfrac{x}{20} &=& 1 \\\\ x &=& 20 \\\\ \mathbf{\overline{AE}} &=& \mathbf{20} \\ \hline \end{array}\)
