ABCD is a trapezium having AB || DC. Prove that O, the point of intersection of diagonals, divides the two diagonals in the same ratio. Also prove that

#### Solution

We are given *ABCD* is a trapezium with *AB*||*DC*

Consider the triangles *AOB* and *COD* in which

`∠ AOB = ∠COD `

`∠ ABO = ∠ODC ` (ALTERNATIVE ANGLE)

`∠ BAO = ∠DCA ` (ALTERNATIVE ANGLE)

Therefore, ` ∆ODC ∼ ∆ OBA`

`⇒ (AO)/(OC)=(BO)/(DO)=(AB)/(CD)`

`⇒ (AO)/(OC)=(BO)/(DO) `

Hence we have proved that *O*, the point of intersection of diagonals, divides the two diagonals in the same ratio.

We are given *AB* = 3*CD* and we have to prove that `(ar∆ OCD)/(ar∆ OAB)=1/9`

We already have proved that *AOB* and *COD* are similar triangles

So

`(ar∆ OCD)/(ar∆ OAB)= (CD^2)/(AB^2)`

`(ar∆ OCD)/(ar∆ OAB)= (CD^2)/(3CD^2)`

`(ar∆ OCD)/(ar∆ OAB)= 1/9`

Hence, Prove that `(ar∆ OCD)/(ar∆ OAB)= 1/9`