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2) Prove that the midpoint of the hypotenuse of a right triangle is equidistant from all three vertices of that triangle. (That is, this midpoint is the circumcenter of this triangle.) Midpoint of hyp. = X AX = BX sqrt((a-0)^2 + (0-b)^2) X = ((a+0)/2, (b+0)/2) X = (a/2, b/2) distance from (0,0) to X (a/2, b/2) sqrt((0-a/2)^2 + (0-b/2)^2)

          2) Prove that the midpoint of the hypotenuse of a right triangle is equidistant from all three vertices of that triangle. (That is, this midpoint is the circumcenter of this triangle.)
Midpoint of hyp. = X
AX = BX
sqrt((a-0)^2 + (0-b)^2)
X = ((a+0)/2, (b+0)/2)
X = (a/2, b/2)
distance from (0,0) to X (a/2, b/2)
sqrt((0-a/2)^2 + (0-b/2)^2)

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2) Prove that the midpoint of the hypotenuse of a right triangle is equidistant from all three vertices of that triangle. (That is, this midpoint is the circumcenter of this triangle.)
Midpoint of hyp. = X
AX = BX
sqrt((a-0)^2 + (0-b)^2)
X = ((a+0)/2, (b+0)/2)
X = (a/2, b/2)
distance from (0,0) to X (a/2, b/2)
sqrt((0-a/2)^2 + (0-b/2)^2)

Added by John J.

Geometry A Common Core Curriculum

Ron Larson, Laurie Boswell 1st Edition

Instant Answer

Solved by Expert Jason Taylor-Pestell

06/26/2020

Step 1

The hypotenuse of this triangle is the line segment connecting the points (a,0) and (0,b). The midpoint of this hypotenuse can be found by averaging the x-coordinates and the y-coordinates of the endpoints. In this case, the midpoint is ((a+0)/2, (0+b)/2) = (a/2, Show more…

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Prove that the midpoint of the hypotenuse of a right triangle is equidistant from all three vertices of that triangle. (That is, this midpoint is the circumcenter of this triangle.)

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00:01 All right, so this problem begins with a diagram, as shown here, that shows any right triangle being oriented so that its right angle vertex is at the origin, and the two legs of this right triangle are along the axes, x -axis and y -axis, and the hypotenuse is stretched from one.

00:31 One axis to the other.

00:33 And our goal is to prove that this point d is equidistant from b, c, and a, all three points of the triangle's vertices.

00:47 So to begin with, what we're going to do is, like they said, prove by finding several distances.

00:57 Our first distance is going to be the distance from b to c.

01:02 So the distance from b to c, they actually said the answer already on the problem, but i'm going to show how to do that.

01:09 So b, c, or you could also say the distance of the side length, b, c, is equal to the square root of, and what you do is you subtract the x values and square it, then subtract the y values and square that, and this is the distance formula.

01:32 That you should know from previous sections.

01:37 So the x values are 0 and x.

01:40 And the order doesn't matter because when we square it, a negative will become positive.

01:44 Positive will stay positive.

01:46 So i'll just do 0 minus x.

01:49 And then y minus 0.

01:55 So then that means that this will be negative x squared, which is x squared.

01:59 So square root of x squared plus y squared.

02:06 Which is what they said.

02:08 So the distance from b to c is the square root of x squared plus y squared.

02:13 What that means is that since d is the midpoint, it's half of that.

02:18 So the next thing they told us to do was say, determine what the distance is between b and d and between d and c.

02:29 So we'll say that bd is equal to half of the square root of x squared plus y squared, and dc is also equal to half of the square root of x squared plus y squared.

02:52 Okay.

02:53 So we've just found these two distances, and they are equivalent, which that's because it's the midpoint.

03:00 So that's not any new information.

03:03 But now we have an algebraic expression that shows what their distances are.

03:10 Next, it says in the hint, it said, you need to find the distance between a and d.

03:19 So to do that, the first thing we need to is figure out what the coordinates of d are.

03:25 So the coordinates of d are the midpoint between 0y and x, y.

03:32 Sorry, x is 0.

03:33 So what you do is you add the x values and take half of that and add the y values and take half of that.

03:41 So x plus 0 is x.

03:44 Half of that is x over 2, or half of that is x over 2...

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