```Question 292544
The other tutor's answer is incorrect.

Given triangle ACB is a right triangle and D is the midpoint of AB.
CD = 25

{{{drawing(400,126+2/3,-5,55,-2,17, triangle(0,0,0,14,24,7),
triangle(0,0,24,7,48,0), locate(-1.5,0,C), locate(-1.5,16,A),
locate(48,0,B),locate(23,10,D), locate(-2.5,8,14), locate(12,3.5,25)
)}}}

Draw DE perpendicular to AC

{{{drawing(400,126+2/3,-5,55,-2,17, triangle(0,0,0,14,24,7),
triangle(0,0,24,7,48,0), locate(-1.5,0,C), locate(-1.5,16,A),
locate(48,0,B),locate(23,10,D), locate(-2.5,8,14), green(line(24,7,24,0)),
locate(23,0,E), locate(12,3.5,25)
)}}}

Triangle DEB is similar to triangle ACB.
So since D is the midpoint of AB, DB={{{1/2}}}AB,
So, DE={{{1/2}}}AC = {{{1/2}}}(14) = 7.  So we mark DE
as having length 7:

{{{drawing(400,126+2/3,-5,55,-2,17, triangle(0,0,0,14,24,7),
triangle(0,0,24,7,48,0), locate(-1.5,0,C), locate(-1.5,16,A),
locate(48,0,B),locate(23,10,D), locate(-2.5,8,14), green(line(24,7,24,0)),
locate(23,0,E),locate(22,5,7), locate(12,3.5,25)
)}}}

Next we use the Pythagorean theorem on right triangle DEC to find
the length of CE:

{{{CD^2=CE^2+DE^2}}}
{{{25^2=CE^2+7^2}}}
{{{625=CE^2+49}}}
{{{576=CE^2}}}
{{{sqrt(576)=CE}}}
{{{24=CE}}}

so we mark CE as 24:

{{{drawing(400,126+2/3,-5,55,-2,17, triangle(0,0,0,14,24,7),
triangle(0,0,24,7,48,0), locate(-1.5,0,C), locate(-1.5,16,A),
locate(48,0,B),locate(23,10,D), locate(-2.5,8,14), green(line(24,7,24,0)),
locate(23,0,E),locate(22,5,7), locate(12,3.5,25), locate(11.5,0,24)
)}}}

Since triangle DEB is similar to triangle ACB, and DB is half of AB,
BE is half of BC, which makes BE = CE = 24.  So we label BE as 24 also:

{{{drawing(400,126+2/3,-5,55,-2,17, triangle(0,0,0,14,24,7),
triangle(0,0,24,7,48,0), locate(-1.5,0,C), locate(-1.5,16,A),
locate(48,0,B),locate(23,10,D), locate(-2.5,8,14), green(line(24,7,24,0)),
locate(23,0,E),locate(22,5,7), locate(12,3.5,25), locate(11.5,0,24),
locate(35.5,0,24)  )}}}

So that makes the other leg BC = 24+24 = 48.

Edwin</pre> ```