Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Diagram showing A at (0,3), B at (5,4), C at (4,-1) and E at (2,1)

Show that BE is the perpendicular bisector to AC.

I tried to prove this through Pythagoras, but the answer I got did not prove it was at a right angle, and therefore said it was not the perpendicular bisector. I'm probably making a silly mistake, so any advice would be helpful!

share|cite|improve this question
you can solve this exactly the same way as the previous question you asked: find the equation of the line AC, then the perpendicular bisector. If the points E and B lie on the perpendicular bisector then you have proved the theorem. – anon Sep 18 '10 at 18:26
If you push up all the points and lines by one unit, you can recognize the 45 degree right angled triangle and normal at midpoint. – Narasimham Sep 1 '15 at 11:07
up vote 2 down vote accepted

Consider the slopes of $AE$ and $EB$. $AE$ has a slope of $-1$, whereas $EB$ has a slope of $1$. Two lines on a plane are perpendicular if their slopes multiply to $-1$, so since $-1\ast 1=-1,$ the lines are in fact perpendicular.

share|cite|improve this answer
Thank you so much! – Charlotte Sep 18 '10 at 18:24

A meta-analysis:

The calculation with Pythagoras' theorem should in fact work for the points indicated.

If done correctly it has to be consistent with any correct method.

I hope the result of the Q & A is not to treat the posted solutions as correct (maybe because there are several of them that all support the problem statement, sound authoritative, are posted by high-ranked math.SE users, or you checked the reasoning in detail, or whatever) but simultaneously disregard the original computational evidence that another equally valid principle seems to give a different answer. Tracing back to locate the source of the inconsistency should be worthwhile.

share|cite|improve this answer

To expand on Agusti's point further, here's a strategy that only makes use of Pythagoras (remember that the distance formula for two points is a result of the Pythagorean relation!): You have two adjacent triangles $\triangle AEB$ and $\triangle BEC$; compute the distances of all relevant segments, show that the two triangles are right triangles (thus proving the "perpendicular" part) and then show that $\overline{AE}\cong\overline{EC}$ (thus proving the "bisector" part).

share|cite|improve this answer

You can also use the simple general formula that I mentioned in my prior post, namely

HINT $\;$ The equation is $\rm\;\: 2\ (A-B)\cdot (x,y) \;=\; |A|-|B|\;\;\;$ where $\rm\;\;\; |(a,b)| \ =\ a^2 + b^2$

which, if worked out, yields $\rm\;\: (-8,8)\cdot (x,y) \;=\; \;17 \;- 9\;\;\:$ for $\rm\; A = (4,-1),\;\; B = (0,3)$

which, after simplifying, yields the equation $\rm\;\;\: y \;=\; x-1$

share|cite|improve this answer

I don't understand why do you say "I tried to prove this through Pythagoras, but the answer I got did not prove it was at a right angle", because, to me at least, Pythagoras tells me that it is a right angle.

To see this, you must know that Pythagoras' theorem actually is an "if and only if". That is, if you have a triangle with two sides named $u$ and $v$, then the third one is $u-v$ and Pythagoras says:

$$ u\cdot v = 0 \qquad \Longleftrightarrow \qquad \| u \|^2 + \|v\|^2 = \|u-v\|^2 \ . $$

That is, if sides (cathetuses) $u$ and $v$ are perpendicular to each other, then the sum of their squares is equal to the square of the third side (hypotenuse) $u-v$. But the reciprocal is also true, if the equality on the right hand side of the displayed formula above holds, then $u$ and $v$ are perpendicular to each other.

The proof is easy if you have already seen some Linear Algebra (if not, you may skip what follows, and go for the actual computations for your problem down below). So, the proof:

$$ \|u\|^2+ \|v\|^2 - \|u-v\|^2 = u\cdot u + v\cdot v - (u-v)\cdot (u-v) = -2u\cdot v \ . $$

As for your problem, let's compute the three sides:

$$ \|be\| = \sqrt{3^2 + 3^2} = \sqrt{18} \ , \quad \|ce\|= \sqrt{2^2 + 2^2} = \sqrt{8} \quad \text{and} \quad \|bc\|= \sqrt{1^2 + 5^2} = \sqrt{26} \ . $$

And the quantities appearing in the right hand side of the "extended" Pythagoras' theorem:

$$ \|be\|^2 + \|ce\|^2 = 18 + 8 = 26 \qquad \text{and} \qquad \|bc\|^2 = 26 \ . $$

Hence, $be$ and $ce$ are perpendicular to each other.

share|cite|improve this answer

To show that the segments are perpendicular, consider their slopes. The product of the slopes of perpendicular lines is -1.

To show that BE bisects AC, show that the distance AE is equal to the distance EC.

share|cite|improve this answer

BE is perpendicular to AC, that is correct. You don't need any analytic geometry to see that.

Observe that

AE = EC, so E is the midpoint of AC.

AB = CB, so ABC is isosceles triangle with B being the the vertex opposite the base.

Thus BE is perpendicular to AC, as the line joining the vertex opposite the base (B) to the midpoint of base (E) is perpendicular to the base (AC).

Of course, analytic geometry is probably more useful in similar situations.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.