In triangle $\Delta ABC$,$\frac{DA}{DB} = \frac{CA}{CB}$, $\angle ADB = \angle ACB+90^{\omicron}$, prove that $\frac{AB‧CD}{AC‧BD} = \sqrt{2}$.

Question

I draw an arbitrary $\Delta ABD$ first and get point $E$, then construct $C$ (as intersection of Apollonius circle according to $\Delta ABD$ and circumcircle of $\Delta ABE$). I want to prove that $\angle DCE = 45^{\omicron}$ so I can finish the proof. I have tried analytic geometry but it's too complicated. Please give me some hint or other way to prove, thank you very much.

Answer 1

I solved the problem finally. According to $\frac{DA}{DB} = \frac{CA}{CB}$, construct Apollonius circle G

$\angle DCE = \angle DCB + \angle BCE$
$= \angle FCB - \angle FCD + \angle BCE$
$= \frac{1}{2}\angle ACB - \angle FHD + \angle BCE$
$= \frac{1}{2}\angle ACB - \frac{1}{2}(\angle DAB-\angle DBA) + \angle DAB$
$= \frac{1}{2}(\angle ACB + \angle DAB +\angle DBA)$
$ = 45^{\omicron} $

$\frac{AB \times CD}{AC \times BD}$
$= \frac{sinC}{sinB} \times \frac{CD}{BD}$
$= \frac{BD}{DI} \times \frac{CD}{BD}$
$= \frac{CD}{DI}$
$= \sqrt{2}$