Answer:
See Below.
Step-by-step explanation:
Let the intersection point above E be K, and let the intersection point below E be J.
Since ABCD is a trapezoid:
[tex]\displaystyle AB\parallel DC[/tex]
Therefore:
[tex]\displaystyle \angle AJF\cong \angle DKF[/tex]
And:
[tex]\displaystyle \angle DFJ\cong \angle DFJ[/tex]
So, by AA-Similarity:
[tex]\displaystyle \Delta DFK\sim \Delta AFJ[/tex]
CPSTP (Corresponding Parts of Similar Triangles are Proportional):
[tex]\displaystyle \frac{KD}{JA}=\frac{FK}{FJ}[/tex]
Likewise, do the same to the other side. Since AB is parallel to DC:
[tex]\displaystyle \angle BJF\cong CKF\text{ and } \angle CFJ\cong \angle CFJ[/tex]
Hence:
[tex]\Delta CFK\sim \Delta BFJ[/tex]
By CPSTP:
[tex]\displaystyle \frac{KC}{JB}=\frac{FK}{FJ}[/tex]
Substitution:
[tex]\displaystyle \frac{KD}{JA}=\frac{KC}{JB}[/tex]
Since AB is parallel to DC:
[tex]\displaystyle \angle CDB\cong \angle ABD[/tex]
By vertical angles:
[tex]\angle DEK\cong BEJ[/tex]
Hence, by AA-Similarity:
[tex]\Delta DKE\sim \Delta BJE[/tex]
CPSTP:
[tex]\displaystyle \frac{KD}{JB}=\frac{KE}{JE}[/tex]
Likewise:
[tex]\angle DCE\cong \angle BAC[/tex]
And by vertical angles:
[tex]\displaystyle \angle CEK\cong \angle AEJ[/tex]
Hence:
[tex]\Delta CKE\sim \Delta AJE[/tex]
CPSTP:
[tex]\displaystyle \frac{KC}{JA}=\frac{KE}{JE}[/tex]
Substitution:
[tex]\displaystyle \frac{KD}{JB}=\frac{KC}{JA}[/tex]
Now, we have obtained that:
[tex]\displaystyle \frac{KD}{JA}=\frac{KC}{JB}[/tex]
And:
[tex]\displaystyle \frac{KD}{JB}=\frac{KC}{JA}[/tex]
In the second proportion, multiply both sides by JA:
[tex]\displaystyle \frac{KD\cdot JA}{JB}=KC[/tex]
In the first equation, multiply both sides by JB:
[tex]\displaystyle \frac{KD\cdot JB}{JA}=KC[/tex]
Substitute:
[tex]\displaystyle \frac{KD\cdot JA}{JB}=\frac{KD\cdot JB}{JA}[/tex]
Cross-multiply:
[tex]JB\cdot KD\cdot JB=JA \cdot KD\cdot JA[/tex]
Cancel out KD and simplify:
[tex]JB^2=JA^2[/tex]
Therefore:
[tex]JB=JA[/tex]
And using the first proportion again:
[tex]\displaystyle \frac{KD}{JA}=\frac{KC}{JA}\Rightarrow KD=KC[/tex]
Hence, EF bisects AB and DC, the bases of the trapezoid.
