Similarity of Triangle problem

Given: AD & PS are medians in ΔABC and ΔPQR respectively,
$$\frac{AB}{PQ}=\frac{AD}{PS}=\frac{AC}{PR}$$

To Prove: ΔABC ~ ΔPQR

Figure:

Problem: In ΔABD & ΔPQS or in ΔADC & ΔPSR or in ΔABC & ΔPQR, I have only found that only two sides are proportional but can't figure-out third thing to prove similarity.

Plz help me.

• I would use the cosine law to prove that one of the angles at $A$ equals one of the angles at $P$ Jun 29 '11 at 4:34
• @Ross: Its a 10th standard question and students don't know trignometry, any other way to prove... Jun 29 '11 at 4:40
• Since D is the midpoint of BC and S the midpoint of QR, you know that BD = DC, QS = SR. So BD/QS = DC/SR ... Jun 29 '11 at 4:50
• You have to use (somehow) the fact that BDC and QSR are straight lines ie that the angles at D (and angles at S) are supplementary. It is possible (easy) to create a configuration where the ratios are equal and QS = QR but QSR is not straight - so straightness is essential. Jun 29 '11 at 5:01

Create a point A' in the direction of AD, such that DA'=AD. Then there is a parallelogram ABA'C. So is parallelogram PQP'R.

So, AB/PQ=AA'/PP'=AC/PR.
=>ΔABA' ~ ΔPQP'
Then ∠BAC=∠QPR
=>ΔABC ~ ΔPQR
(Q.E.D.)

• Nice clean solution. Jun 29 '11 at 13:19
• thanx @puresky, very very fine solution. Jun 29 '11 at 13:38
• I've added an image (created with Geogebra; please feel free to remove or replace with better image or whatever. Jun 29 '11 at 13:42
• Thanx @ShreevatsaR. It's nice of you for helping to make it clear. Jun 30 '11 at 5:27

We can use a special case of Stewart's Theorem, which has a non-trigonometric proof recently discussed on Stack Exchange.

I will therefore write out a sketch of a trigonometric proof of the result, in the knowledge that the trigonometry can be eliminated.

To make our lives easier, so we don't have to work with ratios, scale one of the triangles to make $AB=PQ$. The other sides that were in the same ratio then also become equal. So now we want to prove congruence. Then we can scale back to get similarity.

Look at the left triangle, and let $AB=c$, $AC=B$. Let $AD=d$, and let $BD=DC=u$.

Consider $\triangle ADB$, and let $\theta$ be the angle at $D$.

By the Cosine Law, $$c^2=d^2+u^2-2du \cos(\theta).$$

Now in $\triangle ADC$, the angle at $D$ has cosine the negative of $\cos(\theta)$, so again by the Cosine Law, $$b^2=d^2+u^2+2du\cos(\theta).$$

Add, sneakily letting $\theta$ drop out of the picture. We get $$c^2+b^2=2(d^2+u^2).$$

Comment: To get rid of the trigonometry, we can drop a perpendicular to $BC$ from $A$. Then two applications of the Pythagorean Theorem give us the fact that $c^2+b^2=2(d^2+u^2)$, which will turn out to be what we need. (The details of the calculation are done one of the Stewart's Theorem answers referred to earlier.)

In the other triangle, let $v=QS=SR$. Remember that the other sides are equal to the corresponding sides of the first triangle, because of the scaling. Thus $$c^2+b^2=2(d^2+v^2).$$

It follows that $u=v$, and we are finished.

• There is a name for what you got: Appolonius Theorem Jun 29 '11 at 5:14
• @Aryabhata: Thanks for the name. I will stick with the name of the generalization, because of the happy accident that it was recently discussed on this site. Jun 29 '11 at 5:21
• Am I right that there is a much simpler proof based on amWhy's comment? Wouldn't the triangles be similar even if we weren't told $AD/PS$ is equal to those other ratios? Jun 29 '11 at 6:03
• @Gerry Myerson: There are inequality-based proofs that use less machinery. If you want to get rid of the middle term, you certainly need to add something. Jun 29 '11 at 6:58
• @user6312, thanks, but I don't see where a proof based on amWhy's comment uses anything about $AD/PS$. What am I missing? Is there a counterexample? Jun 29 '11 at 13:09