Question

d is a point on side bc of a triangle abc such that angle adc=angle bac. show that ca²=cb. cd​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\bf\huge\underline{Answer}$

We have a ∆ABC and a point D on its side BC such that ∠ADC = ∠BAC.

In ∆BAC and ∆ADC,

Since, ∠BAC = ∠ADC⠀⠀⠀⠀⠀⠀⠀⠀[Given]

and ∠BCA = ∠DCA ⠀⠀⠀⠀⠀⠀⠀[Common]

Therefore, Using AA similarity, we have

⠀⠀⠀⠀⠀⠀⠀∆BAC ~ ∆ADC.

Therefore, Their corresponding sides are proportional.

=> $\dfrac{CA}{CD}$ = $\dfrac{CB}{CA}$

=> CA × CA = CB × CD

=> CA² = CB × CD