Question

Prove that the perpendicular drawn from the vertex of a regular pentagon to the opposite side bisects that side.

Answer 1

All the triangles that we have divided into a regular pentagon are congruent.

This means that their area is also equal of $\left|AB\right|=\left|BC\right|=\left|CD\right|=\left|DE\right|=\left|EA\right|=a$

In general, the area of a regular polygon with $n$ vertical is equal to :

$\Rightarrow$ A$=n.\frac{a.hc}{2}$

From $P_{1}BS.$ The triangle is right angled. we know that the measure of each interior angle of a regular pentagon is equal to $108°.$ This means $P_{1}BS=54°$, because segment $\overline{BS}$ divides an interior angle $\angle ABC$ of a regular pentagon into angles both of equal measures.

Hence, prove.