We have a trapezoid ABCD and the diagonals intersect in point O. AO : OC = m : n
Prove that DO : OB = m : n
I think this discussion will show you how to proceed.
https://math.stackexchange.com/questions/1706649/problem-on-a-trapezoid-having-intersection-of-diagonals
To prove that DO : OB = m : n in the trapezoid ABCD with diagonals intersecting at point O, where AO : OC = m : n, we can use the property of similar triangles.
Let's consider triangle ADO and triangle BCO.
By the intersection property, we have:
AO : OC = AD : BC
Since we're given that AO : OC = m : n, we can write:
m : n = AD : BC
Now, let's consider triangle ADO and triangle BDO. These two triangles share the same base (DO) and have the same height (the distance from O to the line AB, which is the same as the distance from O to the line CD).
The area of a triangle is given by the formula:
Area = (1/2) * base * height
In both triangles, the base is DO, and the height is the same. Therefore, the ratio of their areas is equal to the ratio of their bases:
Area(ADO) : Area(BDO) = DO : DO
But the area of a triangle is proportional to its base, so:
Area(ADO) : Area(BDO) = AD : BD
Since we know that AD : BD = m : n (from the earlier equation), we have:
Area(ADO) : Area(BDO) = m : n
Since the areas of ADO and BDO have the same ratio as the bases AD and BD, we can conclude that DO : OB = m : n.