# Let ABC be a triangle such that angle ACB is acute. Suppose that D is an interior point of the triangle ABC such that ∡ADB=∡ACB+pi/2 and AC cdot BD=AD cdot BC. (a) Find (AB cdot CD)/(AC cdot BD). (b) Show that the tangents at C to the circumcicle of the triangle ACD and the circumcircle of the triangle BCD are perpendicular.

Let ABC be a triangle such that $\mathrm{\angle }ACB$ is acute. Suppose that D is an interior point of the triangle ABC such that $\measuredangle ADB=\measuredangle ACB+\frac{\pi }{2}$ and $AC\cdot BD=AD\cdot BC\phantom{\rule{thinmathspace}{0ex}}.$.
a) Find $\frac{AB\cdot CD}{AC\cdot BD}\phantom{\rule{thinmathspace}{0ex}}.$.
(b) Show that the tangents at C to the circumcicle of the triangle ACD and the circumcircle of the triangle BCD are perpendicular.
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Urijah Hahn
Step 1
Draw the perpendicular to CB and then choose E on it, s.t. $CB=CE$, as in the picture below. Now obviously $\mathrm{\angle }ACE=\mathrm{\angle }ADB$ and also from the condition:
$\frac{AC}{CE}=\frac{AC}{BC}=\frac{AD}{DB}\phantom{\rule{thinmathspace}{0ex}}.$

Step 2
Therefore $\mathrm{△}ACE\sim \mathrm{△}ADB$. So in particular we have $\mathrm{\angle }CAE=\mathrm{\angle }DAB$. Also from the similarity of the triangles we have that $\frac{AC}{AD}=\frac{AE}{AB}\phantom{\rule{thinmathspace}{0ex}}.$.
This gives us that $\mathrm{△}ACD\sim \mathrm{△}ABE$. So using that BCE is a right isosceles triangle we have from $\mathrm{△}ACD\sim \mathrm{△}ABE$ that $CD\cdot AB=EB\cdot AD=\sqrt{2}BC\cdot AD=\sqrt{2}AC\cdot BD$.
Hence the ratio is $\sqrt{2}$.

Ethen Blackwell
Step 1
Let's invert about D eith arbitrary radius r>0. For any point X in the plane let X∗ be the image of X under the inversion. Then, we will rewrite all conditions in terms of A∗, B∗, C∗ and D.
Firstly, $\mathrm{\angle }ACB=\mathrm{\angle }ACD+\mathrm{\angle }BCD=\mathrm{\angle }D{A}^{\ast }{C}^{\ast }+\mathrm{\angle }D{B}^{\ast }{C}^{\ast }$ and $\mathrm{\angle }ADB=\mathrm{\angle }{A}^{\ast }D{B}^{\ast }$, so we have $\mathrm{\angle }ADB=\mathrm{\angle }{A}^{\ast }D{B}^{\ast }$.
Hence, $\mathrm{\angle }{A}^{\ast }{C}^{\ast }{B}^{\ast }=\frac{\pi }{2}$.
Step 2
Secondly, recall that for any points M and N (other than D) we have and , so the second equality can be rewritten as $\frac{{R}^{2}}{D{A}^{\ast }\cdot D{C}^{\ast }}\cdot {A}^{\ast }{C}^{\ast }\cdot \frac{{R}^{2}}{D{B}^{\ast }}=\frac{{R}^{2}}{D{B}^{\ast }\cdot D{C}^{\ast }}\cdot {B}^{\ast }{C}^{\ast }\frac{{R}^{2}}{D{A}^{\ast }},$, or ${A}^{\ast }{C}^{\ast }={B}^{\ast }{C}^{\ast }.$.
Thus, the triangle ${A}^{\ast }{B}^{\ast }{C}^{\ast }$ is isosceles and right-angled with $\mathrm{\angle }{A}^{\ast }{C}^{\ast }{B}^{\ast }=\frac{\pi }{2}$