Diagonals of a trapezoid intersect each other at point E, and the extensions of the legs of the trapezoid intersect each other at point F. Prove that line EF bisects the bases of the trapezoid.

PLEASE HELP, IT’S DUE TOMORROW. 50 POINTS AND BRAINLIEST.

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.

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