Do the altitudes of an isosceles triangle go to the midpoints of the opposite sides?

My triangle ABC is isosceles with points A (0,0), B (a,b), and C (2a,0). My diagram shows that the altitude from A will hit side BC at point N and the altitude from C will hit side AB at point M, but I can't find anywhere in the book if M and N are supposed to be midpoints. Help?

Thanks,
Amy :)

You are dealing with the intersection of the three altitudes, which is the orthocentre.

The third altitude, from B to the x-axis will definitely hit the midpoint of AC.
The others will only hit the midpoint of AB and BC if the triangle is equilateral.

for that to happen AB = BC
√(b^2 + a^2) = 2a
b^2 + a^2 = 4a^2
b = √3a, so point B must be (a,√3a)

Proof:
If N is the midpoint then the product of the slopes of BC and AN must be -1
(their slopes must be negative reciprocals of each other)

N is midpoint of BC = (3a/2,√3a/2)
slope of BC = √3a/-a = -√3
slope of AN = √3a/2 ÷ 3a/2 = √3/3

product = -√3(√3/3) = -1
QED

follow the same steps to show M is the midpoint of AB, for the given equilateral condition.

Hi Amy! I'd be happy to help you understand the concept of altitudes in an isosceles triangle.

In an isosceles triangle, the two sides that are equal in length are called the legs, and the third side is called the base. The altitude of a triangle is the perpendicular distance from a vertex to the opposite side or base.

Now, let's consider your triangle ABC with vertices A (0,0), B (a,b), and C (2a,0). To find the equation of the line containing altitude from vertex A, we first need to find the equation of the line containing side BC. The slope of line BC can be found using the formula:

slope = (change in y)/(change in x)

The change in y when going from C to B is b - 0 = b, and the change in x is (2a - a) = a. Therefore, the slope of line BC is b/a.

Since the altitude is perpendicular to side BC, the slope of the altitude will be the negative reciprocal of the slope of line BC. So, the slope of the altitude is -a/b.

Now, we can use the point-slope form of a linear equation to find the equation of the line containing the altitude from vertex A. The point-slope form is given by:

y - y1 = m(x - x1)

where (x1, y1) is any point on the line, and m is the slope of the line. Plugging in the values (x1, y1) = (0,0) and m = -a/b, we have:

y - 0 = (-a/b)(x - 0)

which simplifies to:

y = (-a/b)x

This equation represents the line containing the altitude from vertex A.

Similarly, you can find the equation of the line containing the altitude from vertex C. The slope of line AB is (b - 0)/(a - 0) = b/a, and the slope of the altitude is the negative reciprocal, -a/b. Using the point-slope form again with the point (x1, y1) = (2a, 0), we have:

y - 0 = (-a/b)(x - 2a)

Simplifying this equation gives us:

y = (-a/b)(x - 2a)

Therefore, the equation represents the line containing the altitude from vertex C.

Now, to determine if the altitudes pass through the midpoints of the opposite sides, we can find the coordinates of the intersection points between the altitudes and the opposite sides.

First, let's find the coordinates of point N, where the altitude from vertex A intersects side BC. To find the x-coordinate of N, we equate the equation of the altitude from vertex A to the equation of side BC:

(-a/b)x = b/a(x - 2a)

Simplifying this equation gives us:

(-a/b)x = (bx - 2ab)/a

Multiplying both sides by -ab gives:

abx = -b^2(x - 2a)

Expanding and simplifying further gives us:

abx = -b^2x + 2ab^2

Combining like terms gives:

(ab + b^2)x = 2ab^2

Dividing both sides by (ab + b^2) gives:

x = 2ab^2/(ab + b^2)

Now, to find the y-coordinate of N, we substitute the value of x into the equation of the altitude from vertex A:

y = (-a/b)(2ab^2/(ab + b^2))

Simplifying this equation gives us:

y = -2a^2b/(ab + b^2)

Therefore, the coordinates of point N are x = 2ab^2/(ab + b^2) and y = -2a^2b/(ab + b^2).

Similarly, you can find the coordinates of point M, where the altitude from vertex C intersects side AB. By equating the equation of the altitude from vertex C to the equation of side AB, you can solve for the coordinates of point M.

If the coordinates of points M and N are midpoints of the respective opposite sides, then the altitudes of the isosceles triangle will indeed go through the midpoints. To verify this, you can calculate the distances between the vertices of the triangle and the midpoints M and N. If these distances are equal, then M and N are indeed the midpoints.

I hope this explanation helps you understand the concept of altitudes in an isosceles triangle and how to determine if the altitudes go through the midpoints of the opposite sides. Let me know if you have any further questions!