# Triangle's elementary geometry problem

• Vahn
In summary, the problem involves finding the ratio BP:PC in a right angled triangle with angle A = 90° and AB = AC, where M is the midpoint of AC and P is a point on BC such that AP is vertical to BM. The solution involves using the intercept theorem and the concept of similar triangles, and the final answer is 2.
Vahn

## Homework Statement

Let $\bigtriangleup$ ABC be a right angled triangle such that $\angle$ 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 BP:PC.

## Homework Equations

Possibly the intercept theorem and others related to the congruency of line segments and triangles similarity.

## The Attempt at a Solution

I first tried to attack the problem using vectors, but my limited knowledge of their rules meant I quickly found a dead-end. I then tried expressing BP as BC-PC, and since BC is the hypotenuse, the ratio is $\frac{AB\sqrt{2}}{PC} - 1$, but could not figure how to express PC as a function of AB.

I then spent some days trying to fiddle with proportions to no avail. My last try involved connecting the points C and H, and then extending the resulting line segment until it crossed AB, and called that point C'. By Ceva's theorem, I found out that $\frac{BP}{PC} = \frac{BC'}{AC'}$ and then, using the intercept theorem, $\frac{BP}{BC'} = \frac{PC}{AC'} = \frac{BC}{AB} = \sqrt{2}$. I can't figure out what to do with that information, so it's another dead end.

What I would do is set up a coordinate system. Take A to be at (0, 0), B at (0, 1), and C at (1, 0). The M is at (1/2, 0) and it is easy to find the equation of line BM and then line AP. Use that to find point P.

I see. So, assuming C = (0,0), A = (1, 0), B = (1, 1), I have:

$BM: y_{BM} = 2 x_{BM} - 1$
$AP: y_{AP} = \frac{- x_{AP} + 1}{2}$
$BC: y_{BC} = x_{BC}$

So $P = (\frac{1}{3}, \frac{1}{3})$ and $BP:PC=2:1$, which is the correct answer :).

However, the 'official' resolution of this problem is the following:

1 - $\bigtriangleup AHM = \bigtriangleup HCM$

2 - $\frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{\bigtriangleup AHC}{\bigtriangleup HPC}$

3 - $\frac{\bigtriangleup ABH}{\bigtriangleup HBP} = \frac{2 \bigtriangleup AHM}{\bigtriangleup HPC} = \frac{1}{2}$

4 - $\frac{BP}{PC} = \frac{\bigtriangleup HBP}{\bigtriangleup HPC} = 2$

I'm completely lost.

You can do it with just geometry. At C draw a line parallel to AP meeting BA extended at W. Then triangle AWC is similar to AMB. This gives AW/AM = AC/AB = 2AM/AB or

AW = 2 AM2/AB

Now BP/PC = BA/AW. Substitute AW from the above equation into this giving

BP/PC =(1/2)(BA/AM)2= (1/2)*4 = 2

## What is the Triangle's elementary geometry problem?

The Triangle's elementary geometry problem is a basic geometric problem that involves finding the area, perimeter, angles, and side lengths of a triangle given certain information. It is often used to introduce students to the concepts of geometry and familiarize them with triangle properties.

## What are the key properties of a triangle?

The key properties of a triangle include having three sides, three angles, and the sum of the interior angles always equaling 180 degrees. The sides of a triangle can also be classified as either equal (equilateral), two equal (isosceles), or all different lengths (scalene).

## What is the Pythagorean theorem and how is it used in triangle problems?

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This theorem is frequently used to find missing side lengths or to prove that a triangle is a right triangle.

## What is the difference between acute, obtuse, and right triangles?

An acute triangle has all angles less than 90 degrees, an obtuse triangle has one angle greater than 90 degrees, and a right triangle has one angle equal to 90 degrees. The classification of a triangle as acute, obtuse, or right can also depend on the lengths of its sides and the Pythagorean theorem.

## How can I use the triangle congruence postulates to solve problems?

The triangle congruence postulates (SSS, SAS, ASA, AAS, and HL) state that if certain elements of two triangles are congruent, then the two triangles are also congruent. This means that all corresponding angles and sides of the triangles are equal in length. These postulates can be used to solve problems involving congruent triangles and to prove that two triangles are congruent.

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