# STEP 2 2009 Q8 (spoilers!)

For the shape of the quadrilateral I obtained that opposite sides of the quadrilateral are equal and parallel. I incorrectly came to the conclusion that it is a square. what is an easy way to determine that it was in fact a parallelogram?

Could you please check the working below:

p = λ(a) + (1-λ)b and q = c + μ(a - c)

CQ × BP = AB × AC

⇒ ⟦μ(a -c)⟧⟦λ(a-b)⟧=⟦b-a⟧⟦c-a⟧
⇒⟦μ⟧ = 1/⟦λ⟧
since λ and μ > 0
μ = 1/λ

IF PQ intersects point D then :

r = OP + t(PQ) = d for a parameter t / OP and PQ are vectors

Using this I obtained :

(c + 1/λ (a -c)) + t(a(1/λ - λ) - b(1-λ) + c(1-1/λ)) = -a + b +c

Equating non - parallel vectors:

t= 1/λ -1 ,
t= -1/1-λ,
t= -1/1-λ ( again)

since ⟦λ⟧ < 1 then there exists a value of t for this range, so PQ always passes through D as these equations are consistent.

The first part (showing $\mu = \frac{1}{\lambda}$) is fine. For the second part, you shouldn't start with "If $PQ$ intersects point $D$ then", since you are effectively assuming the thing you are trying to prove. However, you can fix it by saying "$PQ$ passes through point $D$ if and only if ...", since the condition you are using is both necessary and sufficient. The rest of this part is then fine.