Question

$ABCDE$ is a regular pentagon, diagonal $AD$ divides $\angle CDE$ in two parts, what is the ratio of $\angle ADE$ and $\angle ADC$?

Answer 1

Figure shows the regular pentagon, with diagonal AD divides $\angle CDE$ into two parts.

Interior angle $\theta$ of regular pentagon

$(n-2)\times 180=n\times \theta$ [where n=5]

$(5-2)\times 180=5\times \theta$

$540=5\times \theta$

$\theta ={108}^o$

So, $\angle BCD={108}^o$

Diagonal AD is parallel to side BC

hence,

$\angle ADC=180-108={72}^o$

$\angle ADE=108-\angle ADC=108-72={36}^o$

$\therefore \frac{\angle ADE}{\angle ADC}=\frac{36}{72}=\frac{1}{2}$

Hence the ratio of $\angle ADE:\angle ADC=1:2$