# Do you need to measure and then prove that the distances are equal?

Question:

My solution: Using the mid-point formula, we can easily prove that the coordinates of the point $M$ are $(\frac{a}{2}, \frac{b}{2})$. After this, we can use the distance formula to compute the distance $BM$ & $MA$ as follows (I'm doing so in an equation instead of individually):

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

$\implies (0 + \frac{a}{2})^2 + (b + \frac{b}{2})^2 = (a + \frac{a}{2})^2 + (0 + \frac{b}{2})^2 ... (1)$

$\implies \frac{a^2}{4} + b^2 + \frac{b^2}{4} + b^2 = a^2 + \frac{a^2}{4} + a^2 + \frac{b^2}{4}$

$\implies b^2 = a^2$

$\implies b = a$ (since distance can never be negative)

Putting $b=a$ in $1$, we get $0=0.$ Similarly, we can prove for $C$. Now, is this approach correct or do I need to individually calculate the distances and then show that they are equal?

• You have a mistake in the formula of the distance. For example the distance between $M$ and $B$ is $d(M,B)=\sqrt{\left(0-\frac{a}{2}\right)^2+\left(b-\frac{b}{2}\right)^2}=\sqrt{ \frac{a^2}{4}+\frac{b^2}{4}}$. For the proof, check Roman83's answer – Darío G Apr 26 '16 at 14:34
• Oh! I'm so sorry. And I've seen his answer. That's my question. Instead of finding only $d(M, B)$. What if I write something like $d(M, B) = d(M, A)$ which is equation to $0 = 0$. Hence the distances are equal? Will it be all right? – MathEnthusiast Apr 26 '16 at 14:44
• It is true that $d(M,B)=d(M,A)$ simply because $M$ is the midpoint of $AB$. What you have to prove is that $d(0M)=d(MB)$, but this is easy since you already know that $M$ is the point with coordinates $\left(\frac{a}{2},\frac{b}{2}\right)$. – Darío G Apr 26 '16 at 15:06
• Okay! I got it, thanks so much! – MathEnthusiast Apr 26 '16 at 15:16

The very first line under the radical sign is wrong: it should be $b - \frac{b}{2}$ that gets squares, and similarly for $a - \frac{a}{2}$.
You re hoping to show these two square roots are equal, which you do by squaring them, but that doesn't work. Because even though $-2$ and $2$ have squares that are equal, the numbers themselves are not. YOu need to say something like this:
When you then do out the algebra, you need double-implication (both forward and backward) at each step. Otherwise you're showing that something implies a true statement, which tells you nothing about the original thing. For instance, I could claim that $2 = 1$, and therefore $1 = 2$, and by adding equations, get that $3 = 3$. The truth of that last stemenet tells you nothing about the truth of the first one (which is obviously false!).
$$MA=\sqrt {\left( \frac a2 -0\right)^2+\left(\frac b2 -b \right)^2}=\sqrt{\frac {a^2}4+\frac{b^2}4}$$ Similarly, $$MB=MC=\sqrt{\frac {a^2}4+\frac{b^2}4}$$