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Posted November 28th, 2008 by alpha
in

D is a point on the side BC of a triangle ABC such that angle ADC =angle BAC . Show that CA^2=CB. CD

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answer

On November 29th, 2008 Structure says:
$$\triangle ABC\tilda\triangle DAC$$

as they have a common angle ACB=ACD and

$$\angle BAC=\angle ADC$$

Their sides are proportional

$$\frac{AB}{DA}=\frac{BC}{AC}=\frac{CA}{CD}$$

So

$$CA^2=CD*CB$$

The point D is the intersection between circle by A and B tangent to AC and line BC.
Final relation express power of point C with respect to this circle.

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