# Plane Geometry Question: Distinction

1. Mar 24, 2005

### aek

PQR is an equilateral triangle, and A is a point on QR such that RA = 2 QA. Prove that PA^2 = 7QA^2

If someone can help, i'd really appreciate it. Thanks

2. Mar 24, 2005

### hypermorphism

How have you attempted to prove this statement ?

3. Mar 25, 2005

### Curious3141

HINT : Apply cosine rule on triangle PQA.

4. Mar 25, 2005

### aek

Thanks. . .

Why are you answering my question with a question? Just abit ironic thats all. Thanks for the hint hyper, it was helpful.

5. Mar 25, 2005

### Curious3141

I can't even tell if that's sarcasm. In any case, it was I that gave the hint, not hyper.

The policy of this forum is not to do homework or just solve the problem for the poster unless it is evident some effort or at least some serious thinking has been put into the solution by the poster. So I can understand why hyper posted what he did.

Have you solved the problem yet ? If you're still finding difficulty, apply the Cosine Rule as I suggested and see what you get. If stuck, post here with working, then you'll get help.

6. Mar 25, 2005

### Icebreaker

You can choose any arbitrary equilateral triangle PQR and let a point A on QR such that RA = 2QA. Show arithmatically (with numbers) that PA^2 = 7QA^2. Then proceed to argue that all triangles PQA are congruent, and therefore any proportions wrt its lengths (PA^2 = 7QA^2) must remain constant.

7. Mar 25, 2005

### aek

yeah, it was sarcasm

yep i got the point,
but i too know that you have to use the cosine rule and i've tried and couldn't find a relevant answer. If you want proof that i tried, i could scan my work but you being so trustful that wouldn't be necassary :P

8. Mar 25, 2005

### PBRMEASAP

Here is an alternative but longer way to do it that avoids trig. Let M be the midpoint of RQ. Since PQR is equilateral, you know (it's a theorem) that PM must be perpendicular to RQ. Then you can use Pythagoras's theorem a couple of times to find how the different segments relate to QA.

9. Mar 25, 2005

### Curious3141

$$PQ = QR = 3QA$$

$$PA^2 = PQ^2 + QA^2 - 2(PQ)(QA)\cos 60^o$$

$$PA^2 = 9QA^2 + QA^2 - 2(3QA)(QA)(\frac{1}{2})$$

$$PA^2 = 7QA^2$$

Perhaps now the sarcasm can end.

10. Mar 25, 2005

### aek

perhaps it may. well thanks for the help, even though it took some time but truly, i really do appreciate it.