If D is the mid point of side of side BC of a traingle ABC, P and Q are two points lying respectively on the sides AB and BC such that DP is parallel to QA. Prove that ar(triangle CQP) = 1/2 ar(traingle ABC).
You must have a typo.
Make ABC a 3-4-5 right triangle, and let AQ be the altitude to the hypotenuse.
The areas do not fit your hypothesis.
Thank you
To prove that the area of triangle CQP is half the area of triangle ABC, we will use the property that triangles with the same height and different bases have areas in proportion to their bases.
Let's denote the midpoint D as the intersection point of the median from C to side AB. We are given that DP is parallel to QA, which means triangle ABC and triangle DQA are similar.
Hence, we can write the following proportion:
(CQ / CB) = (QP / BA) = (1 / 2) (since D is the midpoint of BC)
This means that the ratio of the lengths of the sides CQ and CB is 1/2, and the ratio of the lengths of the sides QP and BA is also 1/2.
Now, let's consider the height of both triangles. Since DP is parallel to QA, we can conclude that the height of triangle CQP is equal to the height of triangle ABC.
Using the property stated earlier, the area of triangle CQP is half the area of triangle ABC, since the ratio of their bases (CQ and CB) is 1/2.
Hence, we have proven that ar(triangle CQP) = 1/2 ar(triangle ABC).