Question 1109665: In a quadrilateral PQRS, the sides PQ and SR are parallel, and the diagonal QS bisects angle PQR. Let X be the point of intersection of the diagonals PR and QS. Prove that PX/XR = PQ/QR
Normally I have some sort of start on the question but here I am completely stuck- geometry is my weakest part of maths
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! 
Geometry is mostly about triangles.
You just have to find the key triangles.
There are two obvious congruent angles: XQP and XQR.
Knowing that PQ and SR are parallel,
can you prove that triangles XQP and XSR are similar?
All you need to do is show there are two pairs of congruent angles.
What about XSR and XQP as alternate interior angles?
XRS and XPQ are alternate interior too,
and PXQ and RXS are vertical angles.
If XQP and XSR are similar,
then their corresponding side's lengths are proportional,
so  .
That is not what you needed to prove, but we are close.
Knowing that XSR and XQP are congruent because they are alternate interior angles,
and that XQP is congruent to XQR,
you conclude that XSR and XQR are congruent.
As those two angles are angles of triangle QRS,
we figure that QRS is isosceles,
with congruent angles adjacent to base QS,
and congruent legs SR and QR, so  .
Substituting into , you get
 .
|
|
|
