Explain how corollary 3 follows from theorem 4-1.

Consider the figure.

Here the perpendicular bisects the given triangle. So that both triangles thus formed are congruent using ASA congruency, so that by CPCT, will show the base angles as 90 degree, hence perpendicular on base.

Statement of theorem 4.1 as,” If two sides of an isosceles triangle are congruent, then the angles opposite to those sides are congruent.”

Corollary 3 states that, “The bisector of the vertex angle of an isosceles triangle is perpendicular to the base at its midpoint.

In isosceles triangle, using statement of theorem 4.4, $AB=AC\Rightarrow \angle B=\angle C$.

Now in triangle $\mathrm{\Delta }ABD$and $\mathrm{\Delta }ACD$,

1. $AB=AC\Rightarrow \angle B=\angle C$
2. $m\angle BAD=m\angle CAD$
   

So, by ASA congruency, both the triangles are congruent. Thus, by CPCT,

$m\angle BAD=m\angle CAD$

Again, by definition of linear pair

$\begin{array}{l}m\angle BDA+m\angle CDA=180\\ 2\angle BDA=180\\ \angle BDA=90\end{array}$

It shows that$AD$ is perpendicular on base $BC$.

Thus, by above steps, it is clear that in an isosceles triangle, the bisector of the vertex angle of an isosceles triangle is perpendicular to the base at its midpoint, from corollary 3 that is obtained by the following statement of theorem 4.1 that if two sides of an isosceles triangle are congruent, then the angles opposite to those sides are congruent.”