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: Compute the ratio

Previous step Next step

In plain words

The right isosceles construction supplies the final square-root factor.

\frac{AB\cdot CD}{AC\cdot BD}=\frac{BB'}{CB'}=\sqrt2

Detailed analysis

The second similarity gives CD/AC=BB'/AB', while the first gives AB'/AB=BC/BD. Therefore the requested ratio is BB'/CB'. In the right isosceles triangle BCB', this equals the square root of 2.