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,

localid="1650353047404" $$\begin{gathered} m∠1+m∠2+m∠3=180 \\ 72+72+m∠3=180 \\ m∠3=180−144 \\ =36° \end{gathered}$$

Consider$∠1,∠EBC$and$∠5$such that

localid="1650353072558" $$\begin{gathered} m∠1+m∠EBC+m∠5=180 \\ 72+72+m∠5=180 \\ m∠5=180−144 \\ =36° \end{gathered}$$

Draw a perpendicular angle bisector$\stackrel{¯}{AG}$meet at width="24" height="23" role="math">BE¯ 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,

$$\begin{gathered} ∠BAG=180−\left(∠AGB+∠5\right) \\ =180−90−36 \\ =54° \end{gathered}$$

If$∠BAG=54°$then $∠BAG=∠EAG=54°$.

Therefore $∠4$can be found out by

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

Therefore, the measures of$∠1,∠2,∠3,∠4\text{and}∠5$is$72°,72°,36°,108°\text{and}36°$ respectively.