91Ó°ÊÓ

From the given figure it can be observed that $\stackrel{¯}{BE}âˆ\yen \stackrel{¯}{DF}$and $\stackrel{¯}{BC}$is a transversal then $∠EBC$and $∠2$are the alternate interior angles then,

$m∠2=m∠EBC=72°$.

From the given figure it can be observed that $\stackrel{¯}{BE}âˆ\yen \stackrel{¯}{DF}$and $\stackrel{¯}{BC}$is a transversal then ∠1 and ∠2 are the same side interior angles then,

$m∠1=m∠2=72°$.

Consider $\mathrm{Δ}BCF$, by an angle sum theorem,

$\begin{array}{c}m∠1+m∠2+m∠3=180\\ 72+72+m∠3=180\\ m∠3=180−144\\ =36°\end{array}$

Consider ∠1, $∠EBC$and ∠5 such that

$\begin{array}{c}m∠1+m∠EBC+m∠5=180\end{array}$

$72+72+m∠5=180$

$m∠5=180−144$

$=36°$

Draw a perpendicular angle bisector meet at which bisects ∠4 into two equal angles $∠BAG=∠EAG$.

If in $\mathrm{Δ}AGB$measure of $∠AGB=90°$and $∠5=36°$then measure of $∠BAG$can be found by using an angle theorem,

$∠BAG=180−\left(∠AGB+∠5\right)$

$=180−90−36$

$=54°$

If $∠BAG=54°$ then $∠BAG=∠EAG=54°$.

Therefore ∠4 can be found out by

$\begin{array}{c}m∠4=m∠BAG+m∠EAG\\ =54+54\\ =108°\end{array}$

Therefore, the measures of ∠1, ∠2, ∠3, ∠4 and ∠5 is 72°, 72°, 36°, 108° and 36° respectively.