# 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.

• From $AC+BC=AB$ you get $AB=\vec b-\vec a$ – MattG88 Jan 15 '17 at 13:23
• @MattG88 I am not getting your point. – user404716 Jan 15 '17 at 13:40
• How did you get $AB={\vec b\over 2}$? – MattG88 Jan 15 '17 at 13:43
• I think that you get $BC={\vec b\over2}$ not $AB={\vec b\over2}$ – MattG88 Jan 15 '17 at 14:00
• @MattG i m confused now. Can you help me please. – user404716 Jan 15 '17 at 14:02

## 8 Answers

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.

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.

• can you please explain more? – user404716 Jan 16 '17 at 9:16
• Yes sure, I'll add something later – MattG88 Jan 16 '17 at 12:05
• @MattG88 Sorry, but you only proved the converse: if the middle of the side $OB$ is equidistant, then the triangle is right. – Wolfram Jan 28 '17 at 23:41

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 $B.$

• I want prove using one vertex as origin. – user404716 Jan 15 '17 at 15:12
• In that case can you use the inner product? – DanielWainfleet Jan 15 '17 at 15:20
• I dont know how to use. – user404716 Jan 15 '17 at 15:38
• I've added a second proof in my A. – DanielWainfleet Jan 15 '17 at 16:25

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.

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 $\vec{BC}$).

• 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 $OM=AM=BM$.

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 $|OC|=|BC|\,$, $\;|AC|=|OC|=|BC|$.

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.