I am trying to prove that the following triangles are similar.
Following information is given in this regard:
AB, AC & median AD of triangle ABC are respectively proportional to PQ, PR & median PM of triangle PQR.
AB/PQ=AC/PR=AD/PM=x (given)-----(1)
To prove that these two triangles are similar, I am trying to prove: BC/QR=x (S-S-S similarity condition)
I proceeded like this:
AB+AC+BC=P1, PQ+PR+QR=P2-------------(2)
from (1), these equations become:
xPQ + xPR +BC=P1, PQ+PR+QR=P2------ -(3)
=>BC=P1- x(PQ+PR), QR=P2-(PQ+PR)-----(4)
From (4)
BC/QR= (P1-x(PQ+PR))/(P2-(PQ+PR))-----(5)
In the above method so far I have not been able to use the information:
AD/PM=x--------------------- (6)
How do I proceed now to prove BC/QR=x?
I understand that there can be other methods to prove that the two triangles are similar, but I am particularly interested in proving this using Eqn (3) (and possibly Eqn (6), and some other property). The thing is Eqn (3) must be used and it should not be made redundant information.
Best Answer
The resul proved below is probably overkill, but it can be used to solve the problem.
In $\triangle XYZ$, let $M$ be the midpoint of $YZ$. for simplicity, let $x$, $y$, and $z$ be the lengths of the sides opposite $X$, $Y$, and $Z$ respectively. Let $m$ be the length of the median from $X$. Let $\theta=\angle XMY$, and let $\phi=\angle XMZ$. Note that $\theta+\phi=180^\circ$, and therefore $\cos\phi=-\cos\theta$.
By the Cosine Law applied to $\triangle XMY$, we have $$z^2=(x/2)^2+m^2-2(x/2)m\cos\theta.$$ Similarly, $$y^2=(x/2)^2+m^2-2(x/2)m\cos\phi.$$ Add, using the fact that $\cos\phi=-\cos\theta$. We get $$y^2+z^2=\frac{x^2}{2} +2m^2.$$ So we can get an expression for $m^2$ in terms of $x^2$, $y^2$, and $z^2$.
In particular, for your problem, if the two sides and median of the "small" triangle are proportional to the two sides and median of the "big" triangle, then the remaining two sides are proportional, with the same proportionality constant.
The theorem about the length of the median that was proved above can be proved more "geometrically."