91Ó°ÊÓ

The labelled diagram satisfying the given statement is:

The statement is: If a line perpendicular to$\stackrel{¯}{AB}$ passes through the midpoint of $\stackrel{¯}{AB}$, and segments are drawn from any other point on that line to and , then two congruent triangles are formed.

Consider the midpoint of$\stackrel{¯}{AB}$ be $D$.

Therefore, by using the definition of midpoint it can be said that $AD=BD$.

That implies, $ADâ‰\dots BD$.

Therefore, it is given that $AD=BD$.

As, the line is perpendicular to $AB$.

Therefore, by using the definition of perpendicular lines it can be said that $m∠CDA=m∠CDB=90°$.

Therefore, it is given that $m∠CDA=m∠CDB=90°$.

As the triangles formed are to be proved congruent and the triangles formed are $â–³ADC$and $â–³BDC$.

Therefore, it is to be proved that $â–³ADCâ‰\dots â–³BDC$.

In the triangles$â–³ADC$ and $â–³BDC$, it can be noticed that$CD=CD$ by using the reflexive property.

That implies, $CDâ‰\dots CD$.

Therefore, it can be seen that $AD=BD$, $m∠CDA=m∠CDB=90°$and $CD=CD$.

Therefore, the triangles $â–³ADC$and $â–³BDC$are the congruent triangles by using the SAS postulate.

The proof in two-column form is: