Question

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

Answer 1

CA²=CB.CD
⇒CA/CD=CB/CA
⇒CA/CB=CD/CA
in ΔADC and ΔBAC,
CA/CB=CD/AC and ∠C=∠C[∵common angle]
∴ΔADC is ΔBAC
⇒∠ADC=∠BAC

Answer 2

CA²=CB.CD
⇒CA/CD=CB/CA
⇒CA/CB=CD/CA
in ΔADC and ΔBAC,
CA/CB=CD/AC and ∠C=∠C
∴ΔADC is ΔBAC
⇒∠ADC=∠BAC