91Ó°ÊÓ

Draw scalene $\mathrm{Δ}EFG$with $\stackrel{¯}{EH}$and $\stackrel{¯}{FJ}$as medians. Also $P$is on $\stackrel{→}{EH}$such that $\stackrel{¯}{EH}â‰\dots \stackrel{¯}{HP}$and$Q$is on$\stackrel{¯}{FJ}$ such that $\stackrel{¯}{FJ}â‰\dots \stackrel{¯}{JQ}$.

Since,$∠QJGâ‰\dots ∠FJE$are vertically opposite angles and$\stackrel{¯}{GJ}â‰\dots \stackrel{¯}{EJ}$ as$\stackrel{¯}{FJ}$ is median.

Thus, using given statement $\stackrel{¯}{JQ}â‰\dots \stackrel{¯}{FJ}$,$\mathrm{Δ}GJQâ‰\dots \mathrm{Δ}EJF$ by SAS (Side-Angle-Side) Postulate. As corresponding parts of congruent triangle are congruent implies$\stackrel{¯}{GQ}â‰\dots \stackrel{¯}{EF}$

Since,$∠FHEâ‰\dots ∠GHP$ are vertically opposite angles and$\stackrel{¯}{FH}â‰\dots \stackrel{¯}{GH}$ as$\stackrel{¯}{EH}$ is median.

Thus, using given statement $\stackrel{¯}{EH}â‰\dots \stackrel{¯}{HP}$,$\mathrm{Δ}EHFâ‰\dots \mathrm{Δ}GHP$ by SAS (Side-Angle-Side) Postulate. As corresponding parts of congruent triangle are congruent implies$\stackrel{¯}{EF}â‰\dots \stackrel{¯}{GP}$

Using reflexive property,$\stackrel{¯}{GQ}â‰\dots \stackrel{¯}{EF}$ and$\stackrel{¯}{EF}â‰\dots \stackrel{¯}{GP}$ implies$\stackrel{¯}{GQ}â‰\dots \stackrel{¯}{GP}$

Since,$∠QJGâ‰\dots ∠FJE$ are vertically opposite angles and$\stackrel{¯}{GJ}â‰\dots \stackrel{¯}{EJ}$ as$\stackrel{¯}{FJ}$ is median.

Thus, using given statement $\stackrel{¯}{JQ}â‰\dots \stackrel{¯}{FJ}$,$\mathrm{Δ}GJQâ‰\dots \mathrm{Δ}EJF$ by SAS (Side-Angle-Side) Postulate. As corresponding parts of congruent triangle are congruent implies$∠GQJâ‰\dots ∠EFJ$

As$\stackrel{¯}{FQ}$ is transversal to$\stackrel{¯}{GQ}\text{and}\stackrel{¯}{EF}$ and alternate interior angles are equal then lines$\stackrel{¯}{GQ}\text{and}\stackrel{¯}{EF}$ are parallel

Since,$∠FHEâ‰\dots ∠GHP$ are vertically opposite angles and$\stackrel{¯}{FH}â‰\dots \stackrel{¯}{GH}$ as$\stackrel{¯}{EH}$ is median.

Thus, using given statement $\stackrel{¯}{EH}â‰\dots \stackrel{¯}{HP}$,$\mathrm{Δ}EHFâ‰\dots \mathrm{Δ}GHP$by SAS (Side-Angle-Side) Postulate. As corresponding parts of congruent triangle are congruent implies$∠FEHâ‰\dots ∠GPH$

As $\stackrel{¯}{EP}$is transversal to $\stackrel{¯}{GP}\text{and}\stackrel{¯}{EF}$and alternate interior angles are equal then lines $\stackrel{¯}{GP}\text{and}\stackrel{¯}{EF}$are parallel

Thus, $\stackrel{¯}{GQ}\text{and}\stackrel{¯}{GP}$are both parallel to$\stackrel{¯}{EF}$

From part ‘b.’, $\stackrel{¯}{GQ}âˆ\yen \stackrel{¯}{EF}$and$\stackrel{¯}{GP}âˆ\yen \stackrel{¯}{EF}$

Two lines parallel to same line are also parallel to each other.

So,$\stackrel{¯}{GQ}âˆ\yen \stackrel{¯}{GP}$

As, $\stackrel{¯}{GQ}âˆ\yen \stackrel{¯}{GP}$and both lines have common point $G$. When parallel lines have same common point then lines are coinciding and so there end points are collinear.

Thus, $P,G,\text{and}Q$are collinear.