Question

in triangle ABC point d on side BC is such that angle bac is equal to angle adc prove that ca square is equal to the cd ×cb​

Answer 1

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