Median to the Hypotenuse Formula Proof. Right Triangles.

Use the median to the hypotenuse formula to solve this problem and watch the proof of this formula in step by step instruction.
#geometry #righttriangles #median

[SkinnerRobot]

Another excellent video. Thank you.

I came to this video from your video "Calculate the Area of the Right Triangle. Geometry problem." of 28 Sept 2023.

[hanswust]

You are very didactic and therefore very helpful for students.

This channel will grow with thousands of subscribers worldwide.

[thephilosophyofhorror]

I hope your channel does great ^^ Another proof (equally brief) would be to draw the parallelogram; x is half its diagonal (by Thales theorem, it's in analogy to CN/CB=1/2), so it is half of the other diagonal (CB) = AM= 10.

[Okkk]

If you consider AB as a diameter to a circle with center at M it must pass point C (Thales circle theorem). Therefore x is equal to 10.

[Nikioko]

Easy. 10.

Since ABC is a rectangular triangle, AB is the hypotenuse and thus the diameter of the Thales circle. As M is the midpoint of the hypotenuse and therefore the center of the Thales circle, AM and BM are radii of said circle. And so is CM.

[mauroFsc]

Hola a todos...lo razoné así: ese triángulo es rectángulo, luego se puede inscribir en media circunferencia, luego la mediana es también radio e igual a 10...!!

[mindyourownbastardbu]

A much quicker method. Let AB be a diameter of a circle of radius 10 (IE midpoint of the hypotenuse). Any right angle triangle drawn from AB to meet at C must have point C upon the circumference of that circle, ipso facto MC is a radius of that circle and is equal to half the hypotenuse.
This is the obverse of the rule that two chords drawn to meet at a point on the cumference from either end of the diameter must automatically form a 90° angle.

[じーちゃんねる-vn]

median theorem 20^2=AC^2+BC^2=2(x^2+10^2) ∴x=10

[santiagoarosam]

∆ABC is a right triangle and ∠C=90º → ∆ABC Inscribed in a semicircle with diameter AB and radius AM=10 → Point M is the center of the semicircle and point C belongs to the semicircle → MC is a radius of the semicircle → MC=AM=10=X
Thanks and best regards