Question

please solve it... ​

Answer 1

In triangle BAC and triangle ADC

<BAC=<ADC.... given

<BCA=<ACD.... common angles

triangle BAC~triangle ADC

BA/AD=AC/DC=BC/AC.....c. s. s. t

AC/DC=BC/AC

AC*AC=DC*BC

AC^2=DC*BC

CA^2=CB*CD

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