A two step method

[@mohabatkhanmalak1161]

Wonderfull, I did not know of that obscure theorem relating to a right angle triangle. That is why I am here.🌻

[@MathandEngineering]

Ok do you mind if I prove it?

[@mohabatkhanmalak1161]

@@MathandEngineering Yes please, want to see the proof.

[@MathandEngineering]

@@mohabatkhanmalak1161 ok first draw a Right angled triangle ABC A B C Assume the length of AB to be x and BC y Using Pythagorean theorem AC=hyp=√(x²+y²) If you draw the median from the right angle to the Hyp, AC will be bisected, if we assume the point where the median touches the hypotenuse to be P AP = CP = ½√(x²+y²) now a draw line parallel to AB from point P down to BC and label the point Q. We know that any line that touches 2 sides of a triangle and is parallel to the other side will divide them in the same ratio, so since PQ bisects AC it will also bisect BC, therefore BQ=CQ=½y Now we know also, if a line bisects two sides of a triangle, it's length will be half the length of the other sides so PQ = ½x Now look at triangle BPQ P B Q BP = hyp(BPQ) = median of ABC BQ = ½y PQ = ½x Using the Pythagorean theorem (BP)² = (½x)² + (½y)² = ¼x² + ¼y² = ¼(x² + y²) = (½×½)(x² + y²) √(BP)² = √((½×½)(x² + y²)) BP = ½(x² + y²) Recall BP is the median of ABC AND AC is the hypotenuse of ABC and it's length is √(x² + y²) BP is the median from the right angle, and it's length is ½√(x² + y²) Which is half that of AC this proves the Theorem that the median of a Right angled triangle from the vertex of the right angle is half the length of the hypotenuse I hope you understand the explanation, if you don't. Then please tell me, I'll make a video to make it clearer, thank

[@STEAMerBear]

I wish KZbin allowed us to respond with graphics and videos. As I was reading your proof I found myself thinking it seems circular and I am not sure I “know” some of your assumptions. I watch a lot of KZbin math content, and their policy to prevent us from sharing disrupts effective intellectual dialogue. Too bad there’s not an open math forum where we can pose questions and dig our own rabbit holes. I know of some things that come close, but I hesitate to even mention them lest I get blocked.

[@STEAMerBear]

Good on you for seeing the parallelogram. I was thinking, “It looks like adjoining equilateral triangles,” but didn’t see how to get there. You often use theorems and ideas that are either new to me or so rusty I have only a vague memory of them.

[@MathandEngineering]

Yes you got it, because if you calculate the value of a, it's 8m thereby making the triangles which formed the parallelogram to be equilateral. How did you figure out that the two triangles are equilateral without calculating, I am really impressed. All the time I have to use theorems and rules to know the shapes

[@STEAMerBear]

@@MathandEngineering they just looked equilatral-it was a guess. I draw them often for my geometry classes, but I can’t tell whether they are or are not when they’re off by less than 2-3° (such as a 59-59-62 triangle).