91Ó°ÊÓ

The labelled diagram satisfying the given statement is:

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

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

Consider the two equal sides be $AC$and $BC$.

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

As, segments $MP$and $NQ$are drawn from the midpoints $M$and $N$of the legs $AC$and $BC$of an isosceles triangle perpendicular to the base $AB$.

Therefore, $MPâŠ\yen AB$and $NQâŠ\yen AB$.

Therefore, it is given that $MPâŠ\yen AB$and $NQâŠ\yen AB$.

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

As $MPâŠ\yen AB$and $NQâŠ\yen AB$,therefore $m∠MPA=90°$and $m∠NQB=90°$.

Therefore, $m∠MPA=m∠NQB=90°$

That implies $∠MPAâ‰\dots ∠NQB$

As $M$is the midpoint of $AC$and $N$is the midpoint of $BC$.

Therefore, by using the definition of midpoint, it can be said that $\stackrel{¯}{CM}\mathbf{=}\stackrel{¯}{AM}$and $\stackrel{¯}{CN}\mathbf{=}\stackrel{¯}{BN}$.

By using the segment addition postulate it can be noticed that:

$CA=CM+MA$

As, $\stackrel{¯}{CM}\mathbf{=}\stackrel{¯}{AM}$therefore, it can be noticed that:

$$\begin{gathered} CA=CM+MA \\ =MA+MA \\ =2MA \end{gathered}$$

By using the segment addition postulate it can be noticed that:

$CB=CN+NB$

As, $\stackrel{¯}{CN}\mathbf{=}\stackrel{¯}{BN}$,therefore, it can be noticed that:

$$\begin{gathered} CB=CN+NB \\ =NB+NB \\ =2NB \end{gathered}$$

As, $\stackrel{¯}{CA}â‰\dots \stackrel{¯}{CB}$,therefore, by using the theorem 4-2, it can be noticed that $∠CAB=∠CBA$.

As, $\stackrel{¯}{CA}â‰\dots \stackrel{¯}{CB}$, therefore, it can be noticed that:

$\begin{array}{c}CA=CB\\ 2MA=2NQ\\ MA=NQ\end{array}$

Therefore, $\stackrel{¯}{MA}â‰\dots \stackrel{¯}{NQ}$.

Therefore, in the triangles $â–³MPA$and $â–³NQB$, it can be noticed that role="math" localid="1649935743003" $∠MPAâ‰\dots ∠NQB,∠CAB=∠CBA$and $\stackrel{¯}{MA}â‰\dots \stackrel{¯}{NQ}$.

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

The triangles$â–³MPA$ and$â–³NQB$ are the congruent triangles.

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

The proof in two-column form is: