In given figure, if `angleACB=angleCDA,AC=8cm` and `AD=3 cm`, then find BD.

Explanation

Given, AC=8cm, AD=3 cm and `angleACB=angleCDA`
`angleCDA=90^(@)`
`angleACB=angleCDA+90^(@)`

In right angled `DeltaADC, AC^(2)=AD^(2)+CD^(2)`
`rArr (8)^(2)=(3)^(2)+(CD)^(2)`
`rArr 64-9=CD^(2)`
`rArr CD=sqrt(55)cm`
In `DeltaCDB and DeltaADC, angleBDC=angleADC` [each `90^(@)`]
`angleDBC=angleDCA` [each equal to `90^(@)-angleA`]
`therefore DeltaCDB~DeltaADC`
Then, `(CD)/(BD)=(AD)/(CD)`
`rArr CD^(2)=ADxxBD`
`thereforeBD=(CD^(2))/(AD)=((sqrt(55))^(2))/3=55/3cm`