Q. 164.4( 21 Votes )

# In the given figu

Answer :

In ΔACB and ΔCDB,

ABC = CBD (Common)

ACB = CDB (90°)

So, by AA similarity criterion ΔACB ~ ΔCDB

Similarly, In ΔACB and ΔADC,

ABC = ADC (Common)

ACB = ADC (90°)

So, by AA similarity criterion ΔACB ~ ΔADC

We know that if two triangles are similar then the ratio of their corresponding sides is equal. BC2 = AB×BD ….(i)

And AC2 = AB×AD …..(ii)

Dividing (i) and (ii), we get Hence, proved.

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