# Thread: Hard Problems

1. ## Hard Problems

a) In a quadrilateral ABCD, P and R are the midpoints of AB and CD respectively. Also Q and S are points on the sides BC and DA respectively such that BQ = 2QC and DS = 2SA. Prove that the area of the quadrilateral PQRS equals S/2 where S is the area of the quadrilateral ABCD.

b) In a triangle ABC, AC = 2AB and BC/BA = 3/2. H is the foot of the perpendicular from B to AC. Prove that CH/AH = 21/11.

Ummm for b)

let a = AB; then 2a = AC and 3a/2 = BC
let b = AH; then 2a - b = CH
let h = BH; then h^2 = a^2 - b^2 and h^2 = (3a/2)^2 - (2a - b)^2

that's all i could get. Can anyone help?

2. Originally Posted by xwrathbringerx
a) In a quadrilateral ABCD, P and R are the midpoints of AB and CD respectively. Also Q and S are points on the sides BC and DA respectively such that BQ = 2QC and DS = 2SA. Prove that the area of the quadrilateral PQRS equals S/2 where S is the area of the quadrilateral ABCD.

b) In a triangle ABC, AC = 2AB and BC/BA = 3/2. H is the foot of the perpendicular from B to AC. Prove that CH/AH = 21/11.

Ummm for b)

let a = AB; then 2a = AC and 3a/2 = BC
let b = AH; then 2a - b = CH
let h = BH; then h^2 = a^2 - b^2 and h^2 = (3a/2)^2 - (2a - b)^2

that's all i could get. Can anyone help?
Maybe, next time, you'd want to post hard problems like these individually, or one problem per post, so that those among us who like to answer all problems in one post would not just bypass them---because we could not answer yet all of the problems.

The first problem is really hard if that's all there is to it. ABCD is any quadrilateral? No parallel sides? No given interior angle? Hard.
Just trying to figure how to get the area, S, of ABCD is very hard already.

In the second problem, you are on the right track. Continue with your computations and you should arrive to the conclusion that CH/AH = 21/11 indeed.
That is, (2a-b)/b = 21/11 according to your assumptions.