Question

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

Answer 1

you can easily show that the triangles ABC and ADC are similar (C is common,BAC=ADC)
from the ratio:
ABC
DAC => AC/DC=BC/AC=>AC^2=CD*CB

Answer 2

oh i worked an hour for that & finally got it :)

Answer 3

Thanx so much :)