A property of the midpoint of the hypotenuse in a right triangle 
Prove that in a right angled triangle the mid point of the hypotenuse is equidistant from its vertices.

I have asked similar question but with no satisfactory result.
So I solved it by myself. But I am not sure whether its correct or not. So please help me.
Solution:
Let in triangle $\Delta OAB$ right angled at $A$. Where $O$ is origin and $\vec{a}$ is vector along $OA$ and $\vec{b}$ is vector along $OB$.
Let $C$ be the mid point of hypotenuse $OB$.
We have to prove $OC = BC = AC$.
As $C$ is mid point so $OC = BC =  \frac{\vec{b}}{2}$
$AC =  \frac{\vec{b}}{2} - \vec{a}$
I am not getting the value of AC as $\frac{\vec{b}}{2}$
Sorry because I am using my mobile unable to draw and include image.
Anything wrong please tell me with explanation. Thanks.
 A: Any right triangle can be inscribed into a circle with center $C$ where the hypotenuse $OB$ is the diameter. Now it is straightforward to deduce that $CA$ is a radius, so $CO=CB=CA=r$. 
This is a result of elementary geometry, in fact it is possible demonstrate that in any circle the angle at the centre is twice the corresponding angle at the circumference: $\angle OCB=2\angle OAB$: Euclid's inscribed angle theorem.

Now, if the angle $\angle OCB=180$, then $\angle OAB=90$, $OB$ is a diameter and $\triangle OAB$ is a right triangle.

A: In $\triangle ABC$ let $A'$ be the mid-point of $BC.$  Let $B'$ lie on $AC$ with $A'B'\| AB.$ Then $\angle CAB =\angle CB'A$ because $BA\| A'B'.$ And $\angle CBA=\angle CA'B'$ for the same reason. So triangles $CAB$ and  $CB'A'$ are similar. So $$CB'/CA=CA'/CB=1/2.$$ So $B'$ is the mid-point of $AC.$
Suppose $\angle CAB$ is a right angle. Then so is $\angle CB'A'.$  Since also$AB'=CB',$ this means that $A'$ lies on the right bisector of $CA$. Therefore A' is equidistant from A and C. That is, $$A'A=A'C.$$ Interchanging the letters $B,C$  throughout all of this we also get $$A'A=A'B.$$
In response to comments by the OP I am adding a second proof.
Let $O$ be the origin. Let $\|v\|$ denote the length of a vector $v.$
Let $B,C$ be non-zero vectors with $\angle BOA$ being a right angle. Let $l_1$ be the line through $O,C.$ Let $l_2$ be the line through $O,B.$ Let $l_3$ be the line through $B$ parallel to $l_1.$ We have  $l_1\bot l_2$ and $l_1\| l_3$ so $l_2\bot l_3.$
The vectors $B+C$  and $B-C$ lie on $l_3.$  The distances from $B+C$ to $B$, and from $ B-C$ to $B$ are, respectively, $\|(B+C)-B\|=\|C\|$ and $\| (B-C)-B\|=\|-C\|=\|C\|,$ which are equal distances. 
And $l_2$ contains $B$ with $l_2\bot l_3.$ So $l_2$ is the right bisector of the segment joining $B-C$ to $B+C.$ And $O$ lies on $l_3.$ Therefore the distances from  $B+C$ to $O$ and from  $B-C$ to $O$ are equal. That is $$(\bullet ) \quad \|B+C\|=\|(B+C)-O\|=\|(B-C)-O\|=\|B-C\|.$$
The mid-point $P$ of $B, C$ is $(B+C)/2.$
The distance from $P$ to $B$ is $$\|(B+C)/2-B\|=\|(B-C)/2\|=\|B-C\|/2.$$
The distance from $P$ to $O$ is $$ \|(B+C)/2-O\|=\|(B+C)/2\|=\|B+C\|/2.$$ From $(\bullet )$ we see that $P$ is equidistant from $O$ and from $B.$   
A: A geometric proof is much simpler. Generate a copy ACD of your ABC triangle so as to form a rectangle ABCD. The diagonals AC and QB intesect in their mid points (0), which happens to be the center of the circumscribed circle. Therefore OA=OB=OC=Hypothenuse/2.

A: Here is an image for the question:

As $C$ is the midpoint of $\vec{AB}$, $d$ and $e$ are the midpoints of $\vec{OA}$ and $\vec{AB }$ respectively, and so by symmetry, $|\vec{AC}|=|\vec{OC}|=|\vec{BC}|$.
To see it another way, we can say $\vec{OC}=(d,e)$ with a magnitude of $\sqrt{d^2+e^2}$. $\vec{AC}=(-d,e)$ and therefore has the same magnitude as $\vec{OC}$ (and $\vec{BC}$).
A: Let ABC be any right angled triangle with right angled at C. Let D is the mid point of [AB].
Take C as the origin $\vec{CA} = \vec{a}$
The $\vec{a} \cdot \vec{b} = 0$
Also position vector of A = $\vec{a}$
And position vector of B = $\vec{b}$
Position vector of D = $\frac{\text{P.V of A + P.V of B}}{2} = \frac{\vec{a} + \vec{b}}{2}$
$CD^2 = |\vec{CD}|^2 = |\text{P.V of D}|^2$
= $\left| \frac{\vec{a} + \vec{b}}{2} \right|^2 = \left( \frac{\vec{a} + \vec{b}}{2} \right)^2$
= $\frac 14 (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b})$
= $\frac 14 (a^2 + b^2 + 2 \vec{a} \cdot \vec{b})$
= $\frac 14 (a^2 + b^2)$
As $ \vec{a} \cdot \vec{b} = 0$
Also $AD^2 = |\vec{AD}|^2$
= $|\text{P.V of D - P.V of A}|^2$
= $\left| \frac{\vec{a} + \vec{b}}{2} - \vec{a} \right|^2$
= $\left( \frac{\vec{b} - \vec{a}}{2} \right)^2$
= $\frac 14 (\vec{b} - \vec{a}) \cdot (\vec{b} - \vec{a})$
= $\frac 14 (b^2 + a^2 - 2 \vec{b} \cdot \vec{a})$
= $\frac 14 (a^2 + b^2)$
As $ \vec{a} \cdot \vec{b} = 0$
Hence $CD^2 = AD^2$
$\to |CD| = |AD|$
Also |BD| = |AD|
Then |AD| = |BD| = |CD|
Hence proved.
A: 
Suppose the sign "?" is one of the signs "<", "=", ">" in the expression:
$$AM=CM\ \ ? \ \ BM$$
If you already know that the longer side, the larger opposite angle, you can use this to see that "?" is the same in the next two expressions:
$$\gamma\ ?\ \alpha$$
$$\delta\ ?\ \beta$$
We can add them:
$$\gamma+\delta\ ?\ \alpha+\beta$$
Using $\alpha+\beta+\gamma+\delta=180^o$, it follows that:
$$\gamma+\delta\ ?\ 90^o$$
Thus $BM$ is greater than $AM$ and $CM$ iff the angle $B$ is acute, equal iff the angle $B$ is right and less than iff the angle is obtuse.
A: *

*Assume WLOG that the right-angled triangle has points $O(0,0), A(a,0), B(0,b)$ with $AB$ being the hypotenuse.   

*The midpoint of the hypotenuse $M(\frac a2, \frac b2$).   

*Since the length of $AB$ is $\ell=\sqrt{a^2+b^2}$, $AM=BM=\frac 12 \ell$. 

*The length $OM$ is $\sqrt{\left(\frac a2\right)^2+\left(\frac b2\right)^2}=\frac 12 \sqrt{a^2+b^2}=\frac 12 \ell$  

*Hence $OM=AM=BM$. 

A: 
As $C$ is mid point so $OC = BC =  \frac{\vec{b}}{2}$

Not that it matters to the following, but the second vector is reversed: $\;\overrightarrow{OC} = \overrightarrow{CB} = \frac{\vec b}{2}\,$.

$AC =  \frac{\vec{b}}{2} - \vec{a}$

Let for brevity $\,\vec c = \overrightarrow{OC} = \frac{\vec b}{2}\,$, then $\overrightarrow{AC} = \vec c - \vec a\,$

I am not getting the value of AC as $\frac{\vec{b}}{2}$

That equality is obviously not true as vectors, but what we actually need to get is $|\vec{AC}| = \big|\frac{\vec{b}}{2}\big|=|\vec c|\,$.
In order to get the magnitude of $|\vec{AC}|$ and prove that equality, we have to use the fact that $\triangle OAB$ is a right triangle. The premise $OA \perp AB$ translates in vector notation to the dot product being $0\,$: $$\overrightarrow{OA} \cdot \overrightarrow{AB} = 0 \quad\iff\quad \vec a \cdot (\vec b - \vec a) = 0 \quad\iff\quad \vec a \cdot \vec b - |\vec a|^2 = 0 \tag{1}$$
Since $\vec b = 2\,\vec c$ relation $(1)$ can also be written as:
$$2\,\vec a \cdot \vec c - |\vec a|^2 = 0 \tag{2}$$
Using $(2)\,$, it follows that:
$$
\require{cancel}
|\vec{AC}|^2 = \overrightarrow{AC} \cdot \overrightarrow{AC} = (\vec c - \vec a) \cdot (\vec c - \vec a) = |\vec c|^2 \cancel{- 2\,\vec a \cdot \vec c} + \cancel{|\vec a|^2} = |\vec c|^2 \tag{3}
$$
Therefore $|\vec{AC}|^2 = |\vec{OC}|^2\,$, so $|AC| = |OC|\,$ and, since $|OC|=|BC|\,$, $\;|AC|=|OC|=|BC|$.
A: Let the triangle be ABC with angle B = 90 and AC as hypotenuse. Join B to midpoint of AC. Midpoint of AC is called D.
Let angle C = x. So, angle A = 90 - x.
Let angle DBC = y. So, angle DBA = 90 - y
Now, according to Sine rule of triangle, in ∆DBC:
BD/Sin(C) = DC/Sin(angle DBC)
==> BD/Sin(x) = DC/Sin(y)
==> BD = (DC * Sin(x))/Sin(y) -------------------- (equation 1)
Now, according to Sine rule of triangle, in ∆DBA:
BD/Sin(A) = AD/Sin(angle DBA)
==> BD/Sin(90 - x) = AD/Sin(90 - y)
==> BD = (AD * Cos(x))/Cos(y) -------------------- (equation 2)
We know that AD = DC, So:
BD = (DC * Cos(x))/Cos(y) -------------------- (equation 3)
So, now from equation 1 and equation 3, we get:
(DC * Sin(x))/Sin(y) =  (DC * Cos(x))/Cos(y)
==> Sin(x)/Sin(y) = Cos(x)/Cos(y)
==> Sin(x)Cos(y) = Cos(x)Sin(y)
==> Sin(x)Cos(y) - Cos(x)Sin(y) = 0
==> Sin(x - y) = 0
==> x - y = 0
==> x = y
So, angle C = angle DBC.
So, ∆DBC is isosceles triangle and so,
BD = DC.
So, BD = DC = AD --------- (Proved)
------End of proof--------
