91Ó°ÊÓ

The $\mathrm{Δ}ABC$is:

In $\mathrm{Δ}ABC$, midpoints of $\stackrel{¯}{AB}$, $\stackrel{¯}{AC}$ and$\stackrel{¯}{BC}$ are X, Y and Z respectively. P is the midpoint of$\stackrel{¯}{BZ}$ and Q is the midpoint of $\stackrel{¯}{CZ}$ .

In $\mathrm{Δ}ABC$,

Midpoints of $\stackrel{¯}{AB}$, $\stackrel{¯}{AC}$ are X, Y respectively.

According to the midpoint theorem,

In a triangle, the line segment joining the midpoints of two sides of the triangle is parallel to the third side and is also half of the length of the third side.

So,

$XYâˆ\yen PQ$ …… (i)

Also,

$XY=\frac{1}{2}BC$ …… (ii)

As P is the midpoint of $\stackrel{¯}{BZ}$, $BP=PZ=\frac{1}{2}BZ$.

As Q is the midpoint of $\stackrel{¯}{CZ}$, $ZQ=QC=\frac{1}{2}ZC$.

As midpoint of $\stackrel{¯}{BC}$ is Z, $BZ=ZC$.

$\begin{array}{c}BZ=ZC\\ \frac{1}{2}BZ=\frac{1}{2}ZC\\ PZ=ZQ\end{array}$

Therefore,

$BP=PZ=ZQ=QC$

$PZ+ZQ=BP+QC$

Then,

$PQ=\frac{1}{2}BC$ …… (iii)

From (ii) and (iii),

$XY=PQ$ …… (iv)

From (i) and (iv),

$XY=PQ$, $XYâˆ\yen PQ$.

In a quadrilateral, if one pair of sides is equal and parallel then it is a parallelogram.

Therefore,

$XYQP$is a parallelogram.

So,

$PX=QY$

(Opposite sides of a parallelogram are equal)

Therefore, $PX=QY$(proved).