Submitted by mahmed103 on Sun, 04/07/2019 - 10:20
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.
Re: Quadrilateral
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.
A parallelogram is defined as a quadrilateral where both pairs of opposite sides are parallel. It is easy to show that an equivalent definition is "a quadrilateral where at least one pair of opposite sides are both parallel and equal in length". In order to show that a quadrilateral is a square, you must show that all four sides are equal in length, and that the non-parallel sides are perpendicular.