Question

in the figure, in triangle ABC point D on the side BC is such that angle BAC=angle ADC.​

Answer 1

In ∆ADC and ∆BAC,

angle ADC = angle BAC (given)

angle ACD = angle BCA (common)

therefore, ∆ADC ~ ∆BAC (By AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion .

Therefore, CA/CB = CD/CA

= CA 2 = CB. CD