91Ó°ÊÓ

The labelled diagram satisfying the given statement is:

The statement is: If segments are drawn from the endpoints of the base of an isosceles triangle perpendicular to the opposite legs, then those segments are congruent.

Consider the isosceles triangle be $â–³XYZ$.

Consider the two equal sides be$XY$ and $XZ$.

Therefore, it is given that $\stackrel{¯}{XY}â‰\dots \stackrel{¯}{XZ}$.

As, segments$ZA$and$YB$ are drawn from the endpoints$Y$and$Z$perpendicular to the opposite sides$XY$and $XZ$.

Therefore, $\stackrel{¯}{ZA}âŠ\yen \stackrel{¯}{XY}$and $\stackrel{¯}{YB}âŠ\yen \stackrel{¯}{XZ}$.

Therefore, it is given that $\stackrel{¯}{ZA}âŠ\yen \stackrel{¯}{XY}$and $\stackrel{¯}{YB}âŠ\yen \stackrel{¯}{XZ}$.

It is to be proved that $\stackrel{¯}{ZA}â‰\dots \stackrel{¯}{YB}$.

As, $\stackrel{¯}{ZA}âŠ\yen \stackrel{¯}{XY}$,therefore by using definition of perpendicular lines it can be said that $m∠XAZ=90°$.

As, $\stackrel{¯}{YB}âŠ\yen \stackrel{¯}{XZ}$,therefore by using definition of perpendicular lines it can be said that $m∠XBY=90°$.

Therefore, $∠XBYâ‰\dots ∠XAZ$.

In the triangles$△XBY$ and $△XAZ$, it can be noticed that the angle$∠X$ is common.

Therefore,$∠Xâ‰\dots ∠X$ by using the reflexive property.

Therefore, in the triangles$â–³XBY$ and $â–³XAZ$, it can be noticed that $∠Xâ‰\dots ∠X$,$∠XBYâ‰\dots ∠XAZ$ and $\stackrel{¯}{XY}â‰\dots \stackrel{¯}{XZ}$.

Therefore, the triangles$â–³XBY$ and$â–³XAZ$ are the congruent angles by using the AAS postulate.

The triangles$â–³XBY$ and$â–³XAZ$are the congruent triangles.

Therefore, by using the corresponding parts of congruent triangles it can be said that $\stackrel{¯}{ZA}â‰\dots \stackrel{¯}{YB}$.

The proof in two-column form is: