91Ó°ÊÓ

Since, point$P$is equidistance from $X\text{and}Y$

So,$\stackrel{¯}{PX}â‰\dots \stackrel{¯}{PY}$

By using reflexive property of congruence, a line segment is congruent to itself

So,$\stackrel{¯}{PQ}â‰\dots \stackrel{¯}{PQ}$

As midpoint divides line segment into two congruent line segments

So,$\stackrel{¯}{XQ}â‰\dots \stackrel{¯}{YQ}$

Using step 1, 2, and 3

$\mathrm{Δ}PXQâ‰\dots \mathrm{Δ}PYQ$by SSS (Side-Side-Side) postulate

When two triangles are congruent then there corresponding parts are also congruent

Since,$∠PQX\text{and}∠PQY$ are corresponding parts of congruent triangles

Thus,$∠PQXâ‰\dots ∠PQY$

When two triangles are congruent then there corresponding parts are also congruent

Thus,$∠PQXâ‰\dots ∠PQY;∠XPQâ‰\dots ∠YPQ$

In an isosceles triangle, when median, altitude and the angle bisector of the vertex, all are congruent then median is perpendicular to the base.

So, $\stackrel{↔}{PQ}âŠ\yen XY$

From part (a) $Q$is midpoint of $\stackrel{¯}{XY}$

So, $\stackrel{¯}{PQ}$is median to$\mathrm{Δ}PXY$

From part (c)$∠PQXâ‰\dots ∠PQY$

In an isosceles triangle, when adjacent angles of median are congruent then median is perpendicular bisector.

So, $\stackrel{↔}{PQ}$is perpendicular bisector of$\stackrel{¯}{XY}$