In a triangle ABC, ∠ B A E = ∠ E A D = ∠ D A F = ∠ F A C, and B E : E D : D F : F C = 2 : 1 : 1 : 2..Why is A E : A F = E D : D F? And why B E : E D = B A : A D?Are there any results in ratio or any theorems that I can use to show the equality of the ratio?

Question

In a triangle ABC,   ∠  B  A  E  =  ∠  E  A  D  =  ∠  D  A  F  =  ∠  F  A  C, and   B  E  :  E  D  :  D  F  :  F  C  =  2  :  1  :  1  :  2..Why is   A  E  :  A  F  =  E  D  :  D  F? And why   B  E  :  E  D  =  B  A  :  A  D?Are there any results in ratio or any theorems that I can use to show the equality of the ratio?

suponeriq · Accepted Answer

Explanation:Use the sin rule on   △  A  E  D and   △  A  F  D, which will get you your first result. The same theorem will also apply to your second question.

Levi Rasmussen · Accepted Answer

Step 1Let me first give the proof of the first part. given that   B  E  :  E  D  :  D  F  :  F  C  =  2  :  1  :  1  :  2  ⇒  B  E      /    2  =  E  D      /    1  =  D  F      /    1  =  F  C      /    2  ⇒  B  E  =  F  C  ,  E  D  =  D  F now the for triangle ADE and triangle ADF, we have   E  D  =  D  F, AD is a common line-segment and   A  E  :  A  F  =  1  :  1  =  E  D  :  D  F.Step 2I think the 2nd part of the problem is not correct and can be checked by explicit calculation of length of AB: let   E  D  =  1  ⇒  B  E  =  2. assume   A  D  =  1; now our previous result basically implies ADB is right angle. So   A  B  =  (      3    2    +      1    2        )    1        /    2  =      10    1        /    2  ⇒  A  B  :  A  D  =  (      10    1        /    2  )  :  1 but   B  E  :  E  D  =  2  :  1

In a triangle ABC, angle BAE= angle EAD= angle DAF= angle FAC, and BE:ED:DF:FC=2:1:1:2. Why is AE:AF=ED:DF? And why BE:ED=BA:AD ?

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