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I am trying to prove that the following triangles are similar.

enter image description here

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:


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.

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I do not think a proof will succeed without using some geometric property of the median that goes beyond the fact that it divides a side into two equal parts. –  André Nicolas Nov 27 '12 at 17:57
Other properties can be used but Eqn-3 must be used, as I updated in my question –  gpuguy Nov 27 '12 at 18:03

2 Answers 2

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."

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This is known as Apollonius' theorem –  Daniel Littlewood Nov 27 '12 at 19:06
@DanielLittlewood: Thanks for the helpful link. I do not usually know the standard names for things, which makes searching kind of difficult. –  André Nicolas Nov 27 '12 at 19:14
If I use this formula along with Eqn no 1, then it makes Eqn no.3 redundant. –  gpuguy Nov 28 '12 at 12:52
@gpuguy: True, one uses the scaling factor, say $\sigma$, and we get directly that $\sigma^2 (QR)^2=(BC)^2$. –  André Nicolas Nov 28 '12 at 16:17

You are missing this equation, which connects length of median to the side $c$ with lengths of other sides: $$m_c=\sqrt{\frac{2a^2+2b^2-c^2}{4}}$$

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