So based on this diagram, we want to prove that line E f is parallel toe line segment BC, and it is given that a that a e over e b equals F over FC. So, really, what we’re proving is the converse of the triangle proportionality serum. Um, which states that if the if a line intersects two sides of a triangle and the and it divides those two sides into proportional segments, then that line must be parallel to the third side of the triangle. So to prove this, um, first of all, it will be helpful. Toe actually, rewrite this proportion in another way, rewrite it as e be over a e equals C over a F. And we’re allowed to do that. Um, since they have the same cross products, you’re allowed to just take the reciprocal of both sides of a proportion. So now that we did that little manipulation, we can go over to our two column proof. And first of all, let’s list are givens. So we’ll say, um, he’d be over a e. You know, even though we had to do a little bit of manipulation on the original givens, um, we could just start from here equals FC over f that’s given from there. Um, we’re actually going to rewrite, um, the numerator er’s of this proportion. Um, And what we’re gonna do is we’ll say e b this segment is the same thing as the larger segment, minus a soon will say a B minus a e over a e equals similarly, um, a c minus, um, a over a f and all we did was substitution there, because we know that is a horrible substitution, but well, because we know from the diagram that, um, line segment E b is just the larger segment minus this segments right here. Um, and we can do that for the other side. From there, we can, um, do a little bit of division so we can separate these fractions as a B over a Yeah, minus. Um, e over a year, which is just gonna become one. Anything over itself is one. And that was just pretty much the reverse of, um, combining…

# SOLVED:Prove the Converse of the Triangle Proportionality Theorem. Given: (A E)

Converse of Triangle Proportionality Theorem

Converse of Triangle Proportionality Theorem