Draw $\Delta \text{ABC}$and label the midpoints of role="math" localid="1637831131121" $\overline{\text{AB}},\overline{\text{AC}}$ and$\overline{\text{BC}}$ as X, Y and Z respectively. Let P be the midpoint of$\overline{\text{BZ}}$ and Q be the midpoint of $\overline{\text{CZ}}$ . Prove that $\text{PX}=\text{QY}$ .

The $\Delta ABC$is:

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

In $\Delta ABC$,

Midpoints of $\overline{AB}$, $\overline{AC}$ 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,

$XY\parallel PQ$ …… (i)

Also,

$XY=\frac{1}{2}BC$ …… (ii)

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

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

As midpoint of $\overline{BC}$ 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),

$XY=PQ$, $XY\parallel PQ$.

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).