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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