From the given figure, ABCD is a trapezium in which AB∥DC,
The diagonals AC and BD intersect at O.
So we have to prove that, AO/OC=BO/OD
Consider the △AOB and △COD,
∠AOB=∠COD … [vertically opposite angles]
∠OAB=∠OCD
Therefore, ∠AOB∼△COD
So, OA/OC=OB/OD
Now by using above result we have to find the value of x if OA=3x−19,OB=x−4,OC=x−3 and OD=4.
OA/OC=OB/OD
(3x−19)/(x−3)=(x−4)/4
By cross multiplication we get,
(x−3)(x−4)=4(3x−19)
X2−4x−3x+12=12x−76
X2−7x+12−12x+76=0
X2−19x+88=0
X2−8x−11x+88=0
X(x−8)−11(x−8)=0
(x−8)(x−11)=0
Take x−8=0
X=8
Then, x−11=0
X=11
Therefore, the value of x is 8 and 11.