1. ## Geometric Proofs

Let triangle ABC be a triangle and let D, E, an F be the midpoints of the sides BC, AC, and AB, respectively
a) Prove that triangle EDC is not similar to triangle ABC.
b) Prove that the congruences segment AF congruent to segment ED, segment AE congruent segment FD, and segment BD congruent to segment EF cannot all hold.

2. Originally Posted by sharkman
Let triangle ABC be a triangle and let D, E, an F be the midpoints of the sides BC, AC, and AB, respectively
a) Prove that triangle EDC is not similar to triangle ABC.
b) Prove that the congruences segment AF congruent to segment ED, segment AE congruent segment FD, and segment BD congruent to segment EF cannot all hold.
Hi sharkman,

If I am interpreting your first question correctly, the segment DE is a midsegment of triangle ABC. DE would be parallel to and equal to one-half segment AB. This would, infact, mean that triangle EDC IS SIMILAR to triangle ABC.

3. ## a or b

Originally Posted by masters
Hi sharkman,

If I am interpreting your first question correctly, the segment DE is a midsegment of triangle ABC. DE would be parallel to and equal to one-half segment AB. This would, infact, mean that triangle EDC IS SIMILAR to triangle ABC.
Did you answer a and b or just a?

4. Originally Posted by sharkman
Did you answer a and b or just a?
Actually, I was questioning the fact that you cannot prove that triangle ABC is not similar to triangle EDC. They are similar.

Angle CED is congruent to angle CAB (corresponding angles)
Angle CDE is congruent to angle CBA (corresponding angles)

Therefore, triangle EDC is similar to triangle ABC because of AA similarity property.

You can answer part (b) by what I stated about the midsegment of a triangle. The midsegment of a triangle is the segment that joins the midpoints of two sides. This segment is parallel to the third side and is one-half the length of the third side. This is enough information to show that

ED = 1/2 AB (triangle midsegment theorem)

AF = 1/2 AB (Given)

Therefore, AF = ED

Using similar logic, you can prove AE = FD and BD = EF.

5. ## ok...

Originally Posted by masters
Actually, I was questioning the fact that you cannot prove that triangle ABC is not similar to triangle EDC. They are similar.

Angle CED is congruent to angle CAB (corresponding angles)
Angle CDE is congruent to angle CBA (corresponding angles)

Therefore, triangle EDC is similar to triangle ABC because of AA similarity property.

You can answer part (b) by what I stated about the midsegment of a triangle. The midsegment of a triangle is the segment that joins the midpoints of two sides. This segment is parallel to the third side and is one-half the length of the third side. This is enough information to show that

ED = 1/2 AB (triangle midsegment theorem)

AF = 1/2 AB (Given)

Therefore, AF = ED

Using similar logic, you can prove AE = FD and BD = EF.
Thanks for the break down of part b.