Using the concept of vector equation of straight lines, show the following problem:

Question

Using the concept of vector equation of straight lines, show the following problem:

If P, Q be the mid points of sides AB, CD of a parallelogram ABCD, show that DP and BQ cut diagonal AC at its points of trisection, which are also the points of trisection of DP and BQ respectively.

How to solve it using the concept of vector equation of straight lines, please help.

Answer 1

Let $B=A+V.$ Then $C=D+V .$ And $Q=D+V/2$ is the mid-point of $CD.$

The lines $\{A+x(Q-A):x\in \mathbb R\}$ and $\{D+y(B-D):y\in \mathbb R\}$ intersect where $A+x(Q-A)=D+y(B-D)$.

If $x=2/3$ and $y=1/3$ we have $$A+x(Q-A)=D+y(B-D)\iff$$ $$\iff 3A+3x(Q-A)=3D+3y(B-D)\iff$$ $$\iff 3A+2(Q-A)=3D+(B-D)\iff$$ $$\iff 2(Q-D)=B-A \iff$$ $$\iff 2(V/2)=B-A\iff V=B-A.$$ When $y=1/3$ the point $D+y(B-D)=(2D+B)/3$ trisects the segment $BD$. And when $x=2/3$ the point $A+x(Q-A)=(A+2Q)/3$ trisects the segment $AQ$.

Answer 2

There is a beautiful way to prove by drawing line BQ. But you want to solve it by using the concept of vector.

HINT. Let DP cut AC at X. Then you must prove that $\vec{XA} + \vec{XD} + \vec{XB} = \vec{0}$.

Answer 3

Without less beauty, let be x,y the paralelogram vectors, parametrized as a linear combination growing into the $y$ axes. Then

• $A=0, P=x/2, Q=y+x/2$

he first cross:

• PD=x/2+t(y-x/2),
• AC=t(x+y)

is solved for $t=1/3$, and the second cross:

• BQ=x+t(y+x/2-x),
• AC=t(x+y)

is solved for $t=2/3$, proving the result.