triangles

 

In triangle ADC and triangle ACB

ADC = ACB               (Both 90^o)

DAC is common

ADC  ACB           (By AA corollary)

So, CD/ AD = CB/ CA        ..........(1)        [By cpct] 

In triangle BDC and triangle ACB

BDC = ACB               (Both 90^o)

DBC is common

BDC  ACB           (By AA corollary)

So, BD/ CD = CB/ CA         ..........(2)        [By cpct]  

Multiplying (1) and (2), we get

CB² /  CA² = BD / AD