Difference b/w two methods in vectors.

Question

Prove that in a right angled triangle the mid point of the hyphotenuse is equidistant from its vertices.

My Solution -

In triangle ABC right angled at A. Let D be the mid point of the hyphotenuse BC. We have to show that AD=CD=BD.

CD=BD=1/2BC.

$\vec{AD} = \vec{AB} + \vec{BD}$

$\vec{AD} = \vec{AB} + 1/2\vec{BC}$

$= \vec{AB} + 1/2(\vec{BA} + \vec{AC}$

$= \vec{AB} - 1/2\vec{AB} + 1/2\vec{AC}$

$= 1/2\vec{AB} + 1/2\vec{AC}$

Then on squaring,

$AB^2= 1/4(AB^2 - 2AB.AC + AC^2)$

AB perpendicular to AC so AB.AC = 0.

$AB^2 = 1/4 BC^2$

AB = 1/2 BC

So AD = BD = CD

Hence proved. But i have few questions.

Question 1 -

Can we solve above question using one vertice as origin? If yes please provide solution.

Qusetion 2 -

I have seen many questions some of them solved using one vertice as origin. And some of them not taking one vertice as origin. How to know where to use one vertice as origin vector or not?

Answer 1

Here, you are using pure vector calculus, but in a certain way, you are trying to decompose all your vectors in the "basis" $(\vec{AB},\vec{AC})$.

You could translate your demonstration by choosing $A$ as the origin, and vectors $\vec i$ and $\vec j$ of length $1$ and directions those of $\vec{AB}$ and $\vec{AC}$, and use cartesian coordinates to do the math.

Those methods are completely equivalent.

Answer 2

"Can we solve above question using one vertex as origin? If yes please provide solution."

Yes, of course. Let the distance from A to B be "b" and the distance from A to C be "c". Set up a coordinate system so that A is at the origin, B on the x-axis, and C on the y-axis. (We can do that because A is a right angle.)

Then B is the point (b, 0) and C is the point (0, c). The midpoint of the hypotenuse is (b/2, c/2). The distance from that point to A, (0, 0), is $\sqrt{(b/2- 0)^2+ (c/2- 0)^2}= \sqrt{b^2/4+ c^2/4}= \frac{1}{2}\sqrt{b^2+ c^2}$. Similarly, the distance from that point to B is $\sqrt{(b/2- b)^2+ (c/2- 0)^2}= \sqrt{b^2/4+ c^2/4}= \frac{1}{2}\sqrt{b^2+ c^2}$ and the distance from that point to C is $\sqrt{(b/2- 0)^2+ (c/2- c)^2}= \sqrt{b^2/4+ c^2/4}= \frac{1}{2}\sqrt{b^2+ c^2}$.

"I have seen many questions some of them solved using one vertice as origin. And some of them not taking one vertice as origin. How to know where to use one vertice as origin vector or not?"

You don't "know"- you try it and see if it works. If it doesn't, try something else!

Answer 3

Let $(A;\vec i,\vec j)$ be an orthonormal frame where $\vec{AB}=b\vec i$ and $\vec{AC}=c\vec i$.

Coordinates are : $A(0,0)$, $B(b,0)$, $C(0,c)$ and $D(b/2,c/2)$, and $$AD=\sqrt{(b/2)^2+(c/2)^2}=\frac12\sqrt{b^2+c^2}=BD=CD$$ The demonstration without choosing an origin is exactly the one you gave...