Some ratios in geometry (triangle) problem

Question

I am solving this geometry problem, but my doubt is only on some parts of the problem and I need these results to solve the whole problem.

So here is the whole problem and some of my doubts.

In a triangle $ABC$, $\angle BAE =\angle EAD = \angle DAF =\angle FAC$, and $BE : ED : DF : FC = 2 : 1 : 1 : 2.$

My question: Why is $AE : AF = ED : DF$ ? And why $BE : ED = BA : AD$ ?

Are there any results in ratio or any theorems that I can use to show the equality of the ratio?

Many many thanks for any help! I really appreciate it.

Answer 1

Use the $\sin$ rule on $\triangle AED$ and $\triangle AFD$, which will get you your first result. The same theorem will also apply to your second question.

Answer 2

:let me first give the proof of the first part. given that BE:ED:DF:FC=2:1:1:2 =>BE/2=ED/1=DF/1=FC/2 =>BE=FC , ED=DF now the for triangle ADE and triangle ADF, we have ED=DF, AD is a common line-segment and AE:AF=1:1=ED:DF. I think the 2nd part of the problem is not correct and can be checked by explicit calculation of length of AB: let ED=1=>BE=2. assume AD=1; now our previous result basically implies ADB is right angle. So AB=(3^2+1^2)^1/2=10^1/2=>AB:AD=(10^1/2):1 but BE:ED=2:1