If $D$ is inside an acute triangle $ABC$ s.t. $\angle ADB=\angle ACB+\pi/2$ and $AC\cdot BD=AD\cdot BC$, then find $\frac{AB\cdot CD}{AC\cdot BD}$.

Question

Sources: IMO 1993, Problem 2, and also If $D$ is inside an acute triangle $ABC$ s.t. $\angle ADB=\angle ACB+\pi/2$ and $AC\cdot BD=AD\cdot BC$, then find $\frac{AB\cdot CD}{AC\cdot BD}.$.

Remark. I think this question can remain closed, even if I have provided more context and made some modifications so the question is not quite the same as the old one.

Problem. Let $ABC$ be a triangle such that $\angle ACB$ is acute. Suppose that $D$ is an interior point of the triangle $ABC$ such that $$\measuredangle{ADB}=\measuredangle{ACB}+\frac{\pi}{2}$$ and $$AC \cdot BD=AD\cdot BC\,.$$

(a) Find $$\frac{AB \cdot CD}{AC \cdot BD}\,.$$

(b) Show that the tangents at $C$ to the circumcicle of the triangle $ACD$ and the circumcircle of the triangle $BCD$ are perpendicular.

We should use $\measuredangle{ADB}=\measuredangle{ACB}+\frac{\pi}{2}$ somehow but I don't know how. My first try was to use $\sin$ or $\cos$ law but they didn't work. Any hints?

Solution Sketch of Part (b).

It is easy to show that the tangents at $D$ to the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal to one another by angle chasing. Then, by symmetry, the tangents at $C$ to the circumcircles of the triangles $ACD$ and $BCD$ are also orthogonal to one another.

Attempt of Part (a).

Although the assumption is that $ABC$ is an acute angle, if the answer is a constant number, then by continuity, the answer is the same if we suppose $\angle ACB=\dfrac{\pi}{2}$. Therefore, $\angle ADB=\pi$, so $D$ lies on $AB$. Let $x:=AC$, $y:=BC$, and $z:=AB$. If $w:=BD$, then the condition $AC\cdot BD=AD\cdot BC$ means $$xw=y(z-w)\,.$$

Hence, $$w=\frac{yz}{x+y}\,.$$ Consequently, $BD=\dfrac{yz}{x+y}$ and $$AD=AC-BD=z-\dfrac{yz}{x+y}=\dfrac{xz}{x+y}\,.$$ By Stewart's Theorem, $$CA^2\cdot BD-CD^2\cdot AB+CB^2\cdot AD-BD\cdot AB\cdot AD=0\,.$$ Thus, $$x^2\cdot\left(\frac{yz}{x+y}\right)-CD^2\cdot z+y^2\cdot \left(\frac{xz}{x+y}\right)-\left(\frac{yz}{x+y}\right)\cdot z\cdot\left(\frac{xz}{x+y}\right)=0\,.$$ This shows that $$CD=\frac{\sqrt{xy\big((x+y)^2-z^2\big)}}{x+y}=\frac{\sqrt{xy\big(x^2+2xy+y^2-z^2\big)}}{x+y}\,.$$ By the Pythagorean Theorem, $x^2+y^2=z^2$, so $$CD=\frac{\sqrt{xy\big(2xy+(x^2+y^2-z^2)\big)}}{x+y}=\frac{\sqrt{2}xy}{x+y}\,.$$ We then get $$\frac{AB\cdot CD}{AC\cdot BD}=\frac{z\cdot\left(\frac{\sqrt{2}xy}{x+y}\right)}{x\cdot\left(\frac{yz}{x+y}\right)}=\sqrt{2}\,.$$ How to solve the problem without assuming $\angle ACB\neq \dfrac{\pi}{2}$?

Postscript. The assumption that $ABC$ is acute in the original problem seems irrelevant. As long as $\angle ACB$ is non-obtuse, the same conclusion holds.

Answer 1

Draw the perpendicular to $CB$ and then choose $E$ on it, s.t. $CB = CE$, as in the picture below. Now obviously $\angle ACE = \angle ADB$ and also from the condition:

$$\frac{AC}{CE} = \frac{AC}{BC} = \frac{AD}{DB}\,.$$

Therefore $\triangle ACE \sim \triangle ADB$. So in particular we have $\angle CAE = \angle DAB$. Also from the similarity of the triangles we have that $$\dfrac{AC}{AD} = \dfrac{AE}{AB}\,.$$ This gives us that $\triangle ACD \sim \triangle ABE$. So using that $BCE$ is a right isosceles triangle we have from $\triangle ACD \sim \triangle ABE$ that

$$CD \cdot AB = EB \cdot AD = \sqrt{2} BC \cdot AD = \sqrt{2} AC \cdot BD$$

Hence the ratio is $\sqrt{2}$.

Answer 2

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, $\angle ACB=\angle ACD+\angle BCD=\angle DA^{*}C^{*}+\angle DB^{*}C^{*}$ and $\angle ADB=\angle A^{*}DB^{*}$, so we have $$ \angle DA^{*}C^{*}+\angle DB^{*}C^{*}=\angle A^{*}DB^{*}. $$ Hence, $\angle A^{*}C^{*}B^{*}=\frac{\pi}{2}$.

Secondly, recall that for any points $M$ and $N$ (other than $D$) we have $$ M^{*}N^{*}=\frac{R^2}{DM\cdot DN}\cdot MN~\text{and}~DM^{*}=\frac{R^2}{DM}, $$ so the second equality can be rewritten as $$ \frac{R^2}{DA^{*}\cdot DC^{*}}\cdot A^{*}C^{*}\cdot\frac{R^2}{DB^{*}}=\frac{R^2}{DB^{*}\cdot DC^{*}}\cdot B^{*}C^{*}\frac{R^2}{DA^{*}}, $$ or $$ A^{*}C^{*}=B^{*}C^{*}. $$ Thus, the triangle $A^{*}B^{*}C^{*}$ is isosceles and right-angled with $\angle A^{*}C^{*}B^{*}=\frac{\pi}{2}$.

Part (a):

Let's compute the fraction $\frac{AB\cdot CD}{AC\cdot BD}$ (in the similar way as above): $$ \frac{AB\cdot CD}{AC\cdot BD}=\left(\frac{R^2}{DA^{*}\cdot DB^{*}}\cdot A^{*}B^{*}\cdot\frac{R^2}{DC^{*}}\right):\left(\frac{R^2}{DA^{*}\cdot DC^{*}}\cdot A^{*}C^{*}\cdot\frac{R^2}{DB^{*}}\right)=\frac{A^{*}B^{*}}{A^{*}C^{*}}=\sqrt{2}. $$

Part (b):

Note that images of circumcircles of triangles $ACD$ and $BCD$ are lines $A^{*}C^{*}$ and $B^{*}C^{*}$, respectively. Images unfer the inversion of tangents to these circles at $C$ are circles $\omega_a$ and $\omega_b$ which are passing through $D$ and tangent lines $A^{*}C^{*}$ and $B^{*}C^{*}$ at point $C^{*}$. Since $A^{*}C^{*}\perp B^{*}C^{*}$ the circles $\omega_a$ and $\omega_b$ are orthogonal, so are preimages of $\omega_a$ and $\omega_b$.