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I was doing a problem and I ran into this triangle:

enter image description here

The problem itself is not that important for this question, but rather the explanation of the problem:

We can see that the upper line bisects two of the sides of the triangle on the left. This tells us that the line is parallel to the line that forms the base of the triangle.

However, I can picture line p being able to bisect the triangle into two 6.4 sides and not give a set of parallel lines. Is this a theorem I can research or am I missing something to reach the conclusion a line bisector creates a set of parallel lines?

It also explained

A line parallel to one side of a triangle divides the other two proportionally.

Is this also another theorem?

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  • $\begingroup$ It is the old old Thalès theorem. $\endgroup$ – hamam_Abdallah May 24 '17 at 22:09
  • $\begingroup$ @Salahamam_Fatima I though the Thales theorem was "The diameter of a circle always subtends a right angle to any point on the circle" $\endgroup$ – Pablo May 24 '17 at 22:12
  • $\begingroup$ This sounded so odd to me, I just had to check it out... Turns out english-speakers now call it the intercept theorem: en.wikipedia.org/wiki/Intercept_theorem I've personally always called that the Thalès theorem. I've never given a name to the property you mentioned about right angles and diameters. $\endgroup$ – N.Bach May 25 '17 at 12:09
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enter image description here

Notice the red triangle and the blue triangle are similar as the two sides of the blue triangle (6.4 and 6.0) are half of the sides of the red triangle (2x6.4 and 2x60) and they share an angle.

So the green angles are congruent.

So the lines are parallel.

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There are several proofs depending on how 'deep' you want to go.

One is that line $A$ perpendicular to a line $B$ which perpendicular to another line $C$ is parallel to $C$, and therefore preserves angles along yet another line $D$, which creates similar triangles.

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  • $\begingroup$ I don't understand how it ties into this example, how can I use perpendicular lines to proof my problem? $\endgroup$ – Pablo May 24 '17 at 23:46
  • $\begingroup$ construct the altitude to the triangle from the base parallel, all parallel lines to it ratio both the other sides into x:y $\endgroup$ – JonMark Perry May 24 '17 at 23:55
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To bisect the triangle, you have to bisect the two sides separately. Only one side has a bisector of length 6.4. The other side has a bisector of length 6.0.

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  • $\begingroup$ is there a name to this property? $\endgroup$ – Pablo May 24 '17 at 23:44
  • $\begingroup$ Not that I know of. $\endgroup$ – marty cohen May 24 '17 at 23:46
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The answer to what I was looking for is called the triangle midsegment theorem. The theorem states:

In a triangle, the segment joining the midpoints of any two sides will be parallel to the third side and half its length

For instance:

enter image description here

So if AD = CD, and AE = EB

Then DE = 1/2 CB

And DE and CD are parallel.

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