`a)`

Vì `ABCD` là hình thoi nên `BC////AF; CD////AE`

Áp dụng định lý Thales trong `triangleEAF` có:

`+` Với `BC////AF`, ta có: `(EB)/(BA)=(EC)/(CF) \ (1)`

`+` Với `CD////AE`, ta có: `(AD)/(DF)=(EC)/(CF) \ (2)`

Từ `(1)` và `(2)=> (EB)/(BA)=(AD)/(DF)` (đpcm)

`b)`

Vì `ABCD` là hình thoi có `hat{A}=60^o` nên `AB=BD=AD`

`=>` `(EB)/(BD)=(BD)/(DF)`

Ta có:

`hat{EBD}=hat{EBA}+hat{ABD}=180^o - hat{ABC}+60^o =180^o -120^o + 60^o=120^o`

`hat{BDF}=hat{FDA}+hat{ADB}=180^o - hat{ADC}+60^o = 180^o - 120^o + 60^o=120^o`

`=> hat{EBD}=hat{BDF}=120^o`

Xét `triangleEBD` và `triangleBDF` có:

`(EB)/(BD)=(BD)/(DF)` (chứng minh trên)

`hat{EBD}=hat{BDF}=120^o`

`=> triangleEBD` $\backsim$  `triangleBDF \ (c.g.c)`

`c)`

Trong `triangleBDI` có:

`hat{BID}=180^o - (hat{IDB}+hat{DBI})`

Vì `triangleEBD` $\backsim$ `triangleBDF` nên `hat{BED}=hat{DBF}`

Trong `triangleEBD` có:

`hat{BED}+hat{EDB}+hat{EBD}=180^o`

`=> hat{BED}+hat{EDB}+ 120^o=180^o`

`=> hat{BED}+hat{EDB}=60^o`

`=> hat{DBF}+hat{EDB}=60^o`

Xét `triangleBDI` có:

`hat{BID}=180^o - (hat{DBI}+hat{IDB})=180^o-(hat{DBF}+hat{EDB})=180^o-60^o=120^o`

Vậy `hat{BID}=120^o` (đpcm)

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