a triangle and a parallelogram are drawn on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram and Triangles drawn on the same base and between the same parallel lines are equal in areaIn  △ABC, D is the mid-point of  AB.P is any point on  BC.CQ∥PD meets  AB in  Q and  ar(△BPQ)  =  y ⋅ ar(△ABC). Then the value of  y is?Select an answer

The problem states that triangle BPQ and triangle ABC are drawn on the same base and between the same parallels. According to the property of triangles, the area of triangle BPQ is half the area of parallelogram BQCP.

Since D is the midpoint of AB, triangle BPD is similar to triangle BQC (by AA similarity as ∠B is common and ∠PBD = ∠QBC because they are alternate angles as PD is parallel to CQ).

Since the triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding heights. But in this case, since D is the midpoint of AB, the height of triangle BPD is half the height of triangle BQC. Therefore, the area of triangle BPD is one-fourth the area of triangle BQC.

Now, triangle BPQ is composed of triangle BPD and triangle DPQ. Since triangle DPQ is identical to triangle BPD (as D is the midpoint of AB), the area of triangle BPQ is twice the area of triangle BPD.

Therefore, the area of triangle BPQ is half the area of triangle BQC, which is the same as saying that the area of triangle BPQ is y times the area of triangle ABC.

So, y = 1/2.