triangles

Asked by suchet_r | 5th Dec, 2009, 10:15: PM

Expert Answer:

 

In triangle ADC and triangle ACB

ADC = ACB               (Both 90o)

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 90o)

DBC is common

BDC  ACB           (By AA corollary)

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

Multiplying (1) and (2), we get

CB/  CA2 = BD / AD

Answered by  | 8th Dec, 2009, 04:04: PM

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