A regular m-sided polygon and a regular n-sided polygon are joined together. Part of the polygons are shown below. $\angle BCD : \angle DCG = 5 : 3$ Mike made some calculations and claimed that $m = 6$ and $n = 10$. Determine if Mike's claim is accurate. Justify your answer. Answer Triangle ABC is an equilateral triangle. AC = CD. $\angle ADF = 20^\circ$. AD bisects $\angle CAF$. Show that AF is parallel to CD. State your reasons. Calculate reflex $\angle BAF$. Answer ___ ° Calculate $\angle FDE$. Answer ___ °

1. Identify the relevant polygon angles as exterior angles

For a regular polygon, each exterior angle is
$\frac{360^\circ}{\text{number of sides}}.$
Here, $\angle BCD$ and $\angle DCG$ are the exterior angles of the regular $m$-sided and regular $n$-sided polygons respectively.

2. Calculate the exterior angles using Mike's claimed values

If $m=6$,
$\angle BCD=\frac{360^\circ}{6}=60^\circ.$
If $n=10$,
$\angle DCG=\frac{360^\circ}{10}=36^\circ.$

3. Compare the calculated angle ratio with the given ratio

Therefore,
$\angle BCD: \angle DCG=60:36.$
Simplifying,
$60:36=5:3.$

4. Conclude whether Mike's claim is accurate

The calculated ratio is the same as the given ratio $5:3$.

Hence Mike's claim is accurate.

1. Find the angle at C in triangle ACD

Since $riangle ABC$ is equilateral,
$\angle ACB=60^\circ.$
Since $B, C, D$ lie on a straight line,
$\angle ACD=180^\circ-60^\circ=120^\circ.$

2. Use the isosceles triangle and angle bisector to find angle CAF

Since $AC=CD$, $riangle ACD$ is isosceles.
$\angle CAD=\angle CDA=\frac{180^\circ-120^\circ}{2}=30^\circ.$
Since $AD$ bisects $\angle CAF$,
$\angle CAF=2\times 30^\circ=60^\circ.$

3. Conclude that AF is parallel to CD

Now
$\angle CAF=60^\circ=\angle ACB.$
These are alternate angles, with $AC$ as the transversal and $CB$ lying on the same straight line as $CD$.

Therefore,
$AF\parallel CD.$

1. Find the smaller angle $\angle BAF$

Since $\triangle ABC$ is equilateral,
$\angle BAC=60^\circ.$
From part (i),
$\angle CAF=60^\circ.$
Hence the smaller angle
$\angle BAF=\angle BAC+\angle CAF=60^\circ+60^\circ=120^\circ.$

2. Calculate the reflex angle

Reflex $\angle BAF$ is the angle around the outside:
$360^\circ-120^\circ=240^\circ.$

1. Find \(\angle FDE\) using the angle at D and the given \(20^\circ\) angle

From the earlier result, in $\triangle ACD$, $AC=CD$ and $\angle ACD=120^\circ$, so
$\angle ADC=30^\circ.$
Since $C,D,E$ are on a straight line,
$\angle ADE=180^\circ-30^\circ=150^\circ.$
Given $\angle ADF=20^\circ$,
$\angle FDE=150^\circ-20^\circ=130^\circ.$