Question

Given: Rectangle $\text{QRST}$; $▱\text{RKST}$;

Prove: $\Delta \text{QSK}$is isosceles.

Short answer

$\Delta \text{QSK}$is isosceles.

Step-by-step solution

Step 1. State the concept used.

Diagonals of rectangles are equal and bisect each other.

The opposite sides of a parallelogram are equal.

Step 2. State the proof.

Construct a rectangle $QRST$and parallelogram $▱RKST$.

Now join $QS$to form triangle $\Delta QSK$.

It is to prove that $\Delta QSK$is isosceles which can be proved by showing any two sides of the triangle are of equal length.

In rectangle $QRST$, $TR$and $SQ$are the diagonals of the rectangle.

As the diagonals of the rectangle are of equal length, it can be said that $SQ=RT$.

Now in the parallelogram $RKST$, $RT$and $SK$are parallel sides of the parallelogram, and thus $RT=KS$.

Therefore, $SQ=KS$.

Thus, in $\Delta QSK$, two sides are equal $\left(SQ=KS\right)$, so we can say that $\Delta QSK$ is an isosceles triangle.

Step 3. State the conclusion.

Therefore, $\Delta QSK$is isosceles (proved).