Question

Write a coordinate proof for the following statement: Any triangle ABC formed so that vertex C is on the perpendicular bisector of AB is an isosceles triangle.

Answer 1

Answer:

Answer is contained in explanation.

Step-by-step explanation:

Description of visual:

I started with the first picture. This is a picture of triangle ABC.

Now I'm going to draw a line segment from vertex C such that it is  a perpendicular bisector of AB.

Proof:

CM is a perpendicular bisectors of AB is a given.

From this we can concluded by definition of perpendicular angles that angle AMC and angle BMC are right angles.

Since angles AMC and BMC are right angles, then they are congruent to each other.

By the definition of bisector and since CM bisects AB, then AM is congruent to MB.

By the reflexive property, we have that CM is congruent to CM.

We can conclude the two triangles, triangle CMA and CMB, are congruent by SAS Postulate.

Since triangles CMA and CMB are congruent, we can conclude that their corresponding parts are congruent.

Since their corresponding parts are congruent, then we now know that side CA and side CB are congruent.

Since two sides of the triangle ABC are congruent to each other, namely side CA and side CB, then the triangle ABC is an isosceles triangle.

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Setup for coordinate geometry proof:

M is the midpoint of AB since CM is a bisector of AB.

Since M is the midpoint of AB, then M is located at the coordinates $(\frac{0+b}{2},\frac{0+0}{2})=(\frac{b}{2},0)$.

We found this point such that the length AM is equal to the length MB.

That is, the distance between A and M is the same as the distance between M and B.

Let's check.

$AM=\sqrt{(\frac{b}{2}-0)^2+(0-0)^2}$

$AM=\sqrt{(\frac{b}{2})^2+0}$

$AM=\sqrt{\frac{b^2}{4}}$

$AM=\frac{\sqrt{b^2}}{\sqrt{4}}$

$AM=\frac{b}{2}$

$MB=\sqrt{(b-\frac{b}{2})^2+(0-0)^2}$

$MB=\sqrt{(\frac{b}{2})^2+0}$

$MB=\sqrt{\frac{b^2}{4}}$

$MB=\frac{\sqrt{b^2}}{\sqrt{4}}$

$MB=\frac{b}{2}$

We have confirmed that AM=MB.

(Based on the picture, we could have taken a slightly easier route to calculate the distance between M and A, then the distance between B and M. They are both a horizontal distance. So $MB=b-\frac{b}{2}=\frac{b}{2}$ where as $AM=\frac{b}{2}-0=\frac{b}{2}$.)

Now we also want to assume that the line segment CM is perpendicular to AB. I have drawn the base of the triangle on the x-axis so a vertical line would be perpendicular to it. Also this would make point $C=(c,d)=(\frac{b}{2},d)$. The $y$-coordinate is $d$ because we don't know how high above the $x$-axis the point $C$ is.

If we show CA=CB, then we have shown triangle ABC is an isosceles.

Coordinate Geometry Proof:

We want to finally show that the sides CB and CA of triangle ABC are congruent. We will do this using distance formula.

That is we want to show the distance between (b/2,d) and (0,0) is the same as (b/2,d) and (b,0).

$CB=\sqrt{(b-\frac{b}{2})^2+(0-d)^2}$

$CB=\sqrt{(\frac{b}{2})^2+(-d)^2}$

$CB=\sqrt{\frac{b^2}{4}+d^2}$

$CA=\sqrt{(\frac{b}{2}-0)^2+(d-0)^2}$

$CA=\sqrt{(\frac{b}{2})^2+d^2$

$CA=\sqrt{\frac{b^2}{4}+d^2$

Thus, CA=CB. Since CA=CB, then the triangle is an isosceles.

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