# Thread: A problem on triangles (area, ratios)

1. ## A problem on triangles (area, ratios)

Let Triangle ABC be a right triangle such that <A = 90*, AB = AC and let M be the midpoint of the side AC. Take the point P on the side BC so that AP is vertical to BM. Let H be the intersection point of AP and BM.

Find the ratio between the 2 triangles ABH and AHM
Find the ratio of BP : PC

So far the -BH^2 + HM^2 = 3 is the only equation I was able to formulate :/ Help please.

2. ## Answer for part 1?

Can i just say that

"Since the internal angles are equal among the 2 triangles, then the triangles are deemed similar. If so...

Area of BHA = BA = 2AM = 2
Area of AHM AM AM

Since 2AM = BA
"

3. ## problem of triangles

Hi gundanium.
Assume a 6x6 rt triangle.Compare the area ofAHM to ABM and then ABH to ABM. From these two compare the areas of AHM to ABH.

bjh

4. What do you mean compare? Cause I'm having trouble finding the relations of the other points/segments to point H. Other than that H forms a right angle for the 2 triangles BHA and AHM

5. Hi gundanium,
Triangle ABH is similar to AHM . AM = 1/2 ofAB. Area of AHM =1/4 of ABH. Numbers would certainly help for the second part of problem which is not easy.If you try it and repost I will help

6. First off. How are you able to determine that the triangles are similar? I can only see one similarity which is that they are right triangles from point H. But it ends there.

7. Actually, you have all three angles in the triangles being the same.

Let angle BAH be x. Angle ABH then is 90 - x.

The angle HAM is 90-x and the angle AMH is x.

8. Thank you! I did not see that!

I FINALLY SOLVED THE FIRST PART. YES! It involved a lot of finding the relationships between the sides of different triangles to one another. My approach was to find the relationship of Triangle BAM with AHM to get AH. Then from there related Triangle BAH to AHM then. Put them into A=bxh/2 and came out 4:1. Is that the easiest/most practical way to solve it? Seems long.

Moving on to part II. I can't find any parameter/relation of point P to anything...

9. Hi gundanium,

Part2 first stage,
Let AB =6 then BM =3rad5
Let BH=x and MH = 3rad5 -x

x^2 +AH^2 =36
Solve forx and 3rad5-x and then for AH

If you get this then stage 3 will be next

bjh

10. I'm sorry but... and this might be a very stupid question but... what is 3rad5? how did you get that? I'm not so much familiar when using radians.

11. Well, as soon as you prove that triangles ABH and AHM are similar, you say that side AM corresponds to side AH and AM:AH = 1:2

The area is then simply the square, which is 1:4 for triangle ABH:AHM.

2. Oh well, I used some trigonometry for that (sine rule), and it gets quite long, but I get 1:2.25 = 4 : 9 (for PC : AP)

EDIT: To answer this, 3rad5 is $3\sqrt5$

12. Originally Posted by Unknown008
Well, as soon as you prove that triangles ABH and AHM are similar, you say that side AM corresponds to side AH and AM:AH = 1:2

The area is then simply the square, which is 1:4 for triangle ABH:AHM.

2. Oh well, I used some trigonometry for that (sine rule), and it gets quite long, but I get 1:2.25 = 4 : 9 (for PC : AP)

EDIT: To answer this, 3rad5 is $3\sqrt5$

AM:AH = 1:2? Really? Cause I came up with Sqrt(5)/2
And "the area is simply the square"? How did squaring a side ratio equivalent to area ratio?

OH 3rad5 HAHA NOOB Thanks man

13. Originally Posted by bjhopper
Hi gundanium,

Part2 first stage,
Let AB =6 then BM =3rad5
Let BH=x and MH = 3rad5 -x

x^2 +AH^2 =36
Solve forx and 3rad5-x and then for AH

If you get this then stage 3 will be next

bjh

Yes, i have MH and BH already from my solution when solving the first part. But I don't see how point P will come in.

14. Oops, sorry, I actually meant AM : AB = 1 : 2, my bad

Also, the 'conversion' from length to area is the square of the ratio.

That is, the ratio of length is 1:2, the ratio of area will then be 1^2:2^2 = 1:4

15. REALLY? WOW. I HAD NO IDEA. Is there a derivation for this? I had no idea that to be true. Nor did that exist

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