Triangle $ABC$ is right angled at $A$. The circle with center $A$ and radius $AB$ cuts $BC$ and $AC$ internally at $D$ and $E$ respectively. If $BD = 20$ and $DC = 16$, determine $AC^2$.
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Tags: trigonometry, Pythagorean Theorem, geometry, power of a point
scorpius119
28.01.2006 01:36
Let $AC=a$ and $AB=r$. By Pythagorean Theorem, $a^2+r^2=36^2$.
Extending $\overline{CA}$ to intersect the circle again at $X$, $CX=a+r$. So by Power of a Point,
$CE*CX=CD*CB\Rightarrow (a+r)(a-r)=a^2-r^2=16*36$.
$a^2=\frac{(a^2+r^2)+(a^2-r^2)}{2}=\frac{36(16+36)}{2}=36*26=31^2-5^2=\boxed{936}$.
94337sk
16.01.2015 14:26
use pythagorous theorem on ABC. cosine rule on ADB and ACD.
Virgil Nicula
16.01.2015 19:04
The point $E$ is in addition! Here are two proofs without the power of a point w.r.t. a circle.
Proof 1. I"ll use the Stewart's relation: $20\cdot b^2+16\cdot c^2=36\cdot c^2+20\cdot 16\cdot 36\iff$ $20\cdot b^2=20\cdot c^2+20\cdot 16\cdot 36\iff$
$b^2=c^2+16\cdot 36\implies$ $\left\{\begin{array}{cc}
b^2+c^2=36\cdot 36\\\\
b^2-c^2=16\cdot 36\end{array}\right\|\implies$ $\left\{\begin{array}{cc}
b^2=26\cdot 36\\\\
c=6\sqrt{10}\end{array}\right\|\implies$ $\left\{\begin{array}{cc}
b=6\sqrt{26}\cdot 36\\\\
c^2=10\cdot 36\end{array}\right\|\ .$
Proof 2. Let $K\in BC$ so that $AK\perp BC$ , i.e. $AB=AD\iff KB=KD=10$ . If denote $AK=x$ , then $KB=10$ ,
$KC=26$ and $\left\{\begin{array}{c}
b^2=x^2+26^2\\\\
c^2=x^2+10^2\end{array}\right\|\implies$ $\left\{\begin{array}{cc}
b^2+c^2=36\cdot 36\\\\
b^2-c^2=16\cdot 36\end{array}\right\|\implies$ $\left\{\begin{array}{cc}
b^2=26\cdot 36\\\\
c=6\sqrt{10}\end{array}\right\|\implies$ $\left\{\begin{array}{cc}
b=6\sqrt{26}\cdot 36\\\\
c^2=10\cdot 36\end{array}\right\|\ .$
Remark. You can use directly the equivalence $XY\perp AB\iff XA^2-XB^2=YA^2-YB^2$ .