Answer:
Explanation:
To prove that a quadrilateral ABCD is a trapezium when the diagonals intersect at a point O such that AO/BO=CO/DO, we can use the fact that opposite sides of a trapezium are parallel.
First, let's draw a diagram to represent the given information:
[asy]
pair A, B, C, D, O;
A = (0,0);
B = (2,0);
C = (3,2);
D = (1,2);
O = intersection point(A-C, B--D);
draw(A--B--C--D--cycle);
draw(A-C);
draw(B--D);
label("A", A, W);
label("B", B, E);
label("C", C, E);
label("D", D, W);
label("O", O, N);
label("$\theta_1$", (B+O)/2, NE);
label("$\theta_2$", (O+D)/2, NE);
label("$\theta_3$", (O+C)/2, NW);
label("$\theta_4$", (A+O)/2, NW);
[/asy]
Since the diagonals intersect at point O, the angles formed by the diagonals at O are supplementary (that is, they add up to 180 degrees). In other words, we have:
θ1 + θ2 = 180 degrees
θ3 + θ4 = 180 degrees
Since AO/BO=CO/DO, the triangles AOB and COD are similar. This means that the ratios of their sides are equal. We can write this as:
AO/BO = CO/DO
We can also write this as:
AO/DO = CO/BO
Then, since the angles formed by the diagonals at O are supplementary, we have:
θ1 + θ2 = 180 degrees
θ1 + θ4 = 180 degrees
Subtracting these two equations gives us:
θ2 - θ4 = 0
Since the angles are equal, they are also opposite angles of the quadrilateral ABCD. Therefore, the opposite sides of AB and CD are parallel, which means that ABCD is a trapezium.