Question 1210173
<pre>
I admit I have never heard of Ikleyn's theorem:

If two triangles have a common altitude, then the ratio of their areas
equals the ratio of the bases of these triangles.  

I cannot even find it anywhere online.

The common altitude for the two triangles was not even drawn. It is on the
extension of CA, located outside triangle ABC.

{{{drawing(400,1600/9,-1,8,-1,3,

triangle(0,0,7,0,2.357142857,1.855768722),

green(line(0,0,3.78882,1.28251)),

locate(0,0,B),locate(7,0,C), locate(2.27,2.18,A),
green(locate(3.77,1.6,"D'")),

locate(1.25,1.53,3), locate(3.5,0,7), locate(4.5,1.4,5),

green(locate(2.2,.8,4)), red(line(0,0,2,1),
circle(2,1,.1), locate(2,1.4,P),
line(3.1098313,1.5549156,2.1,1.05),locate(3,2,D)),

triangle(0,0,.85,2.45,7,0), locate(.8,2.3,90^o),locate(.8,2.8,E) 

)}}}

Here, I have drawn in the common altitude BE.  But this is far from 
obvious, especially since the common altitude was not even drawn.

I agree that this is true:

{{{(expr(1/2)("AD'"*BE))/(expr(1/2)(AC*BE))}}}{{{""=""}}}{{{(cross(expr(1/2))("AD'"*cross(BE)))/(cross(expr(1/2))(AC*cross(BE)))}}}{{{""=""}}}{{{"AD'"/(AC)}}}

but I am amazed that it occurred to you.  It certainly would not
have occurred to any ordinary student.  My hat is off to you for 
coming up with that theorem.  Is it included in any geometry book?
As I said above, I cannot find it online.

Thank you for introducing me to Ikleyn's theorem.  My only complaint
is that you assumed that it was something any student would 
immediately recognize, without mentioning anything about the
two triangles having a common altitude.

EDIT LATER:
I finally found Ikleyn's theorem online.  It's called the 
"Same Altitude Principle" on this site:

https://www.youtube.com/watch?v=MQSClqinLUw



Edwin</pre>