# Thread: Prove two areas of triangle by midpoint are equal

1. ## Prove two areas of triangle by midpoint are equal

Here is the question: suppose ABC is a right triangle with the right angle at C. Let M be the midpoint of the segment AB. Prove that the area of triangle AMC is the same as the area of the triangle MCB.

We know that MC=AM=CM. obviously MCB and AMC are isosceles triangles. But this is where I am stuck. I can't really prove that angle A or B are equal, nor the heights of the two triangles the same. Do I use some form of similar triangles? I think that is the process but I don't see how using similar triangles leads me to a congruency. Thank you for the help!!

2. They are similar only when the sides AC and BC are equal. But that's not the case.

Let the length AM = MC = MB = s.

The area of triangle AMC is given by $\displaystyle \dfrac12 s^2 \sin( \theta)$

where theta is the angle AMC.

And angle CMB will be (180 - theta) degrees.

Area of triangle CMB = $\displaystyle \dfrac12 s^2 \sin(180 - \theta)$

Now, recall that $\displaystyle \sin(\theta) = \sin(180 - \theta)$

3. ## area equality

Hello dudeosa,
For any triangle the midpoint of one side and the median drawn from this point to the opposite vertex divides the triangle into equal areas.
Proof
From the above vertex draw a perpendicular to the same side. The area of the triangles created by the median will have the same base lenght and the altitude is common,Remember that the base may need to be extended to erect the perpendicular.If not clear google.Sorry I have no program to make diagrams.

bjh

4. Originally Posted by dudeosu
Here is the question: suppose ABC is a right triangle with the right angle at C. Let M be the midpoint of the segment AB. Prove that the area of triangle AMC is the same as the area of the triangle MCB.

We know that MC=AM=CM. obviously MCB and AMC are isosceles triangles. But this is where I am stuck. I can't really prove that angle A or B are equal, nor the heights of the two triangles the same. Do I use some form of similar triangles? I think that is the process but I don't see how using similar triangles leads me to a congruency. Thank you for the help!!
Recall that in a circle, if you draw a centreline
and pick any point on the circumference not on the edges of that centreline,
then if you join that point to the endpoints of the centreline,
you create a right-angled triangle.
Your point M is the centre of just such a circle.
Hence, the right-angled triangle is formed of 2 isosceles triangles
each with a pair of sides equal to the radius.
Therefore, they have same base lengths and perpendicular heights,
so their areas are equal.

5. Hello, dudeosu!

It's easier than you think . . .

$\displaystyle \text{Suppose }ABC\text{ is a right triangle with }\angle C = 90^o.$
$\displaystyle \text{Let }M\text{ be the midpoint of the segment }AB.$
$\displaystyle \text{Prove that the area of }\Delta AMC\text{ is equal to the area of }\Delta MCB.$
Code:
            C
*
*:\ *
* : \   *
*  :h \     *
*   :   \       *
*    :    \         *
A * - - + - - * - - - - - * B
M

Since $\displaystyle \,M$ is the midpoint of $\displaystyle AB\!:\; AM = MB$
. . The two triangles have equal bases.

The two triangles have the same height, $\displaystyle \,h.$

Therefore, their areas are equal.

Note: This is true for any triangle.
. . . . .A median divides a triangle into two equal areas.

6. Hello again dudeosa,
you were indeed on a right track when you thought similar triangles and congruency.From M draw perpendiculars to the sides. This will give 4 congruent triangles each equal to 1/4 of the total area

bjh

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