Canada
^@ABCD^@ is a parallelogram where ^@P^@ and ^@R^@ are the midpoints of sides ^@BC^@ and ^@DC^@ respectively. If the line ^@PR^@ intersects the diagonal ^@AC^@ at ^@Q^@, prove that ^@AC = 4CQ^@.
Answer:
- Let us draw the image for the situation given in the question.
Also, join ^@ BD ^@ intersecting ^@ AC ^@ at ^@ O ^@.
- It is given that ^@ P ^@ is the mid-point of ^@ BC ^@ and ^@ R ^@ is the mid-point of ^@ DC ^@.
Thus, in triangle ^@CBD^@, by using mid-point theorem ^@PR \parallel BD^@. - As, @^
\begin{aligned}
& PR \parallel BD \\
\implies & PQ \parallel BO \text{ and } QR \parallel OD
\end{aligned}
@^
Now, in triangle ^@BCO^@, we have ^@ PQ \parallel BO ^@ and ^@ P ^@ is the mid point of ^@ BC^@.
By the inverse of mid-point theorem, ^@Q^@ is the midpoint of ^@OC^@. @^ \implies 2 CQ = OC @^ - As the diagonals of a parallelogram bisect each other, ^@ AO = OC ^@.
Thus, @^ \begin{aligned} & AC = AO + OC \\ \implies & AC = 2 OC && [\because \text{ AO = OC]} \\ \implies & AC = 2 \times 2 CQ && [\because \text{ 2 CQ = OC]} \\ \implies & AC = 4 CQ \end{aligned} @^
