Given: ∆ABC, median 𝐴𝐷, AD = 1⁄2 BC
Prove: ∆ABC is a right triangle, and 𝐵𝐶 is its hypotenuse

I will draw the triangle a little "off", so that what we are to prove
does not look quite true.  That way, when creating proofs, we aren't
tempted to assume something to be known that isn't, just because it
looks true.

∠A in the proof, is really ∠BAC, marked with the red arc, and is
the sum of ∠CAD, whose measure is labeled "x", and ∠BAD, whose measure
is labeled "y".




Statements:
1.) 𝐴𝐷 is a median, AD = 1⁄2 BC  | Given
2.) CD = DB                      | Definition of "median" 
3.) CD = AD                      | Given AD = 1⁄2 BC, and CD = 1⁄2 BC  
4.) m∠C = m∠CAD = x              | Renaming measures of angles
5.) DB = AD                      | Things = the same thing are =
6.) m∠B = m∠DAB = 𝑦              | Renaming measures of angles
7.) m∠A = x + y                  | Measures of wholes = sum of measures of
                                   their parts
8.) m∠A + m∠𝐵 + m∠𝐶 = 180o      | Sum of measures of angles of any ∆ is 180o  
9.) (x + y) + x + y = 180o       | Substitution of ='s for ='s
10.) 2x + 2y = 180o              | Addition 
11.) x+y=90o                     | Division of both side by 2
12.) m∠A=m∠CAD+m∠𝐷𝐴𝐵=x+y=90o    | Measures of wholes = sum of measures of
                                   their parts 
13.) ∠A is a right angle         | An angle with measure 90o is a right angle
14.) ∆ABC is a right triangle    | An triangle with a right angle is a right
                                 | triangle
15.) 𝐵𝐶 is its hypotenuse.       | The side opposite the right angle of a
                                 | right triangle is its hypotenuse.

Edwin