Question

in figure in triangle ABC point D on side BC such that angle BAC = angle ADC prove that CA^2 = CB × CD​

Answer 1

$\rm\bf\underline{Question:}$

in figure in triangle ABC point D on side BC such that angle BAC = angle ADC prove that CA^2 = CB × CCD

$\rm\bf\underline{Answer:}$

Given in ΔABC, ∠ADC = ∠BAC

Consider

ΔBAC and ΔADC

∠ADC = ∠BAC (Given)

∠C = ∠C (Common angle)

∴ ΔBAC ~ ΔADC (AA similarity criterion)

AB/AD=CB/CA=CA/CD

Consider CB/CA=CA/CD

CA^2=CB×CD

Answer 2

Given in ΔABC, ∠ADC = ∠BAC

Consider

ΔBAC and ΔADC

∠ADC = ∠BAC (Given)

∠C = ∠C (Common angle)

∴ ΔBAC ~ ΔADC (AA similarity criterion)

AB/AD=CB/CA=CA/CD

Consider CB/CA=CA/CD

CA^2=CB×CD