Question 1082987
<pre><b>
Here is a triangle drawn to scale with AC = 25, and BD = 20.  
But as you see, it cannot have a right angle at B.  The
altitude BD = 20 is just too long to have a right angle at B.

{{{drawing(400,2400/7,-1,6,-1,5,
line(0,-.5,2.3,-.5),locate(2.4,-.3,25),line(2.8,-.5,5,-.5),
triangle(0,0,5,0,3,4), red(line(3,4,3,0),locate(3.1,2,20)),
line(0,-.7,0,-.3),line(5,-.7,5,-.3), locate(-.18,.14,A), locate(5.1,.14,C),
locate(2.93,0,D),locate(2.93,4.3,B)

  )}}}

Here is a triangle also drawn with he same AC = 25 as the
first that has a right angle at B.  However the altitude
is only about 12.2, not 20.

{{{drawing(400,200,-1,6,-1,5,
line(0,-.5,2.3,-.5),locate(2.4,-.3,25),line(2.8,-.5,5,-.5),
triangle(0,0,5,0,3,4), red(line(3,4,3,0),locate(3.1,2,12.2)),
line(0,-.7,0,-.3),line(5,-.7,5,-.3), locate(-.18,.14,A), locate(5.1,.14,C),
locate(2.93,0,D),locate(2.93,4.5,B)  )}}}

This makes us believe that it is impossible to have such a
triangle.

To show that it is impossible, let's suppose that the altitude
BD is actually 20 and angle B is a right angle. Let DC = x, then
AD will = 25-x.


{{{drawing(400,200,-1,6,-1,5,
locate(1.3,.45,25-x),locate(4,.45,x),  

line(0,-.5,2.3,-.5),locate(2.4,-.3,25),line(2.8,-.5,5,-.5),
triangle(0,0,5,0,3,4), red(line(3,4,3,0),locate(3.1,2,20)),
line(0,-.7,0,-.3),line(5,-.7,5,-.3), locate(-.18,.14,A), locate(5.1,.14,C),
locate(2.93,0,D),locate(2.93,4.5,B)  )}}}

Then triangle ADB is similar to triangle BDC, so

{{{AD/"BD"=BD/"DC"}}}

{{{(25-x)/20=20/x}}}

Cross-multiplying:

{{{25x-x^2=400}}}

{{{-x^2+25x-400=0}}}

{{{x^2-25x+400=0}}}

However this quadratic has no real solutions,
because its discriminant is negative:

{{{b^2-4ac=(-25)^2-4(1)(400)=625-1600=-975}}}

Therefore we have proved that there is no such triangle.

Edwin</pre><b>