Problem 2

Let

D

be a point inside acute triangle

ABC

such that

\angle ADB=\angle ACB+\frac{\pi}{2}

and

AC\cdot BD=AD\cdot BC

. (a) Calculate the ratio

\frac{AB\cdot CD}{AC\cdot BD}

. (b) Prove that the tangents at

C

to the circumcircles of

\triangle ACD

and

\triangle BCD

are perpendicular.

Step 3 of 4: Finish part (a)

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In plain words

A Euclidean right triangle supplies the desired ratio before inverting back.

\frac{C'D'}{B'D'}=\sqrt2

Detailed analysis

The right angle and equal legs make triangle B'C'D' right isosceles, so C'D'/B'D' is the square root of 2. Applying the inversion distance formula again identifies this ratio with AB·CD/(AC·BD).