Right Triangle and Circle Theorem

Question

Let $ABC$ be a triangle such that $\angle BAC$ is a right angle. Suppose $D$ is a point lying on $BC$ such that $BD=1$, $DC =3$ and $\angle ADB=60^{\circ}$, find the length of $AC$.

I was told that there is a neat way to solve the question by considering the circumcircle of $\triangle ABC$. Since $\angle BAC=90^{\circ}$, $BC$ must be a diameter. Let $M$ be the midpoint point $BC$, then it's the centre of the circumcircle. But then I couldn't see anything substantial, please helps,

Answer 1

The circumcircle provides a good approach. $|DM|=1$, $|AM|=2$, and $|MC|=2$. Because $\angle ADB = 60^{\circ}$, $\angle ADM = 120^{\circ}$. We may then find $\angle DAM$ using the law of sines:

$$\frac{\sin{\angle DAM}}{|DM|} = \frac{\sin{\angle ADM}}{|AM|} \implies \sin{\angle DAM} = \frac12 \sin{120^{\circ}} = \frac{\sqrt{3}}{4} $$

We may therefore find $\angle AMD = 180^{\circ} - \angle DAM - \angle ADM$ and therefore $\angle AMC = 180^{\circ} - \angle AMD = \angle DAM + \angle ADM$. We may now use the law of cosines to find $|AC|$:

$$\begin{align}|AC|^2 &= |AM|^2+|MC|^2 - 2 |AC| |AM| \cos{\angle AMC}\\&= 2^3 [1- \cos{(120^{\circ}+\angle DAM)}]\\&= 16 \sin^2{\left (60^{\circ}+\frac12 \angle DAM \right )}\\ &= 4 \left (\sin^2{\left (\frac12 \angle DAM \right )}+3 \sin^2{\left (\frac12 \angle DAM \right )} + \sqrt{3} \sin{\angle DAM}\right ) \\ &= 4 \left (\frac{4-\sqrt{13}}{8} + \frac{12+3 \sqrt{13}}{8} + \frac34 \right ) \\ &= 11+\sqrt{13} \end{align}$$

Thus

$$|AC| = \sqrt{11+\sqrt{13}} $$

ADDENDUM

This is even easier when one sees that $\angle ABC = 1/2 \angle AMC$. Also, the previous result I had was correct but not simplified enough.

Answer 2

constuct a circle of radius $2,$ centerd at $0. D = (1,0), B = (2,0), c = (-2,0)$

constuct a vector v at a 60 degree angle with tail at D.

$A = (1 + |v| \cos 60, |v| \sin 60)\\ |A|^2 = 1 + |v| + \frac 14 |v|^2 + \frac34 |v|^2 = 2^2\\ |v|^2 + |v| - 3 = 0$

use the binomial theorm.

$|v| = \frac{-1 + \sqrt{13}}{2}\\ |AC| = \sqrt{(3 + |v| \cos 60)^2 + |v|^2 \sin^2 60}\\ \sqrt{9 + 3|v| + |v|^2)}\\ \sqrt{12 + 2|v| + (-3 + |v| + |v|^2)}\\ \sqrt{12 + 2|v|}\\ \sqrt{11 + \sqrt{13}}\\ $

Answer 3

Let AQ be a chord perpendicular to BC cutting BC at P.

If $PD = x$, then $BP = 1 – x, QP = PA = \sqrt 3 x$ and $AD = 2x$ [$\triangle APD$ is special angled.]

At $P, (1 – x)(x + 3) = (\sqrt 3 x)^2$.

This gives $x = \dfrac {\sqrt {13} - 1}{4}$, after rejecting the negative length

From $\triangle APO$, $(\sqrt 3 x)^2 = 2^2 – (1 + x)^2$ … (*)

From $\triangle APC, AC^2 = (\sqrt 3 x)^2 + (x + 3)^2$ … (#)

Combining (*) and (#), we have $AC^2 = 4x + 12$.

Result follows after putting the value of $x$ back.