Can anyone help me solve these?

Question

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.

Answer 1

line P R divide quadrilateral ABCD with equal area s/2:

A (APRD) = A (PBCR) = S / 2

Sum of the triangle area PRS and PQR is area of the quadrilateral PQRS and is s/2.

It is s/2 not only for  BQ = 2QC and DS = 2SA, it is true for

BQ = N*QC and DS = N*SA

where N=(1/M or M) and M real number !

Triangle area PRS(1/M) equal PQR(M),  sum=S/2  and

                    PQR(1/M) equal PRS(M),  sum=S/2 !

b) In a triangle ABC,  sides are:

   1. AB,

   2. BC=1.5* AB

   3. AC=2* AB

then we know all angles.

and    CH+AH = AC.

COS(angle A) = AH / AB

COS(angle C) = CH / BC = CH / (1.5* AB)

CH / AH = (1.5* AB)*COS(angle C) / (AB * COS(angle A) )

CH / AH = (1.5)*COS(angle C) / (2 - COS(angle C) )

CH / AH = 1.90909...= 21 / 11