Triangles

thedarktiger · 2014-03-14 16:19:12

In triangle ABC, AB = AC, D is the midpoint of \overline{BC}, E is the foot of the perpendicular from D to \overline{AC}, and F is the midpoint of \overline{DE}. Prove that \overline{AF} is perpendicular to \overline{BE}.

thank you

Nehushtan · 2014-03-14 23:29:48

I managed to solve the problem using trigonometry and algebra but its very messy and most likely isnt the solution you want.

We can let A be the origin (0,0) and B be the point (1,0). (Since no lengths are mentioned in the problem we can, without loss of generality, take AB to be of unit length.) If the angle BAC is θ, then

C is the point
D is the point
E is the point
F is the point

A vector parallel to

is

A vector parallel to

is

If you take the scalar product of these two vectors, you should find that it is equal to 0.

Or lets wait for Bob Bundy to come and show us a much simpler solution.

thedarktiger · 2014-03-15 19:55:41

wow... mind BLOWN
thanks

Bob · 2014-03-15 19:57:48

hi

I haven't given up on finding a 'Euclidean' method for this. Brain just waking up so it won't be for a while.

Bob

thedarktiger · 2014-03-17 14:24:50

Thanks... this was in the analytic geometry part...
Good luck!

Bob · 2014-03-18 03:14:39

hi

I think this is it. I've had so many false starts that I'd appreciate a second opinion. The recurring problem is that I kept assuming what I was meant to be proving. It's easy if you do that.

Diagram below.

Let G be the point where BE crosses AF.

I've called the foot of the perpendicular from B to AC, the point H. Note that a circle with AB as diameter goes through H (as AHB is 90) and appears to go through G (which would make AGB 90 as well and that's the problem finished!) So I'll try to avoid using that.

BH is parallel to DE and BC is twice DC so BH is twice DE, and EC = EH.

BE is the hypotenuse of triangle BHE and AF is the hypotenuse of AEF, so I'd like to show angle HBE = angle EAF.

Firstly let angle DAC = x. Then ACD = ADE = 90 - x . So triangles DEC and AED are similar.

So EC/DE = DE/AE .......................................(i)

Tan HBE = EH/BH = EC/2DE

Tan EAF = FE/AE = ½DE/AE = ½EC/DE (using (i) = tan HBE

So HBE = EAF.

If you rotate HBE 90 anticlockwise newBH is parallel to AE and newHE is parallel to FE => newBE is parallel to AF. ie BE and AF are perpendicular.

QED.

Bob

thedarktiger · 2014-03-19 20:51:39

Thanks!
Good proof.

Nehushtan · 2014-03-21 23:52:37

Ah, I knew there had to be a much simpler solution than using coordinates and vectors. Nice work, Bob.

eigenguy · 2014-03-23 06:52:18

A nice proof. Geometric proofs are always lovely. But I have to disagree with Nehushtan's definition of "simple". You came up with that analytic geometry solution within a few hours of the original question. It took Bob days to produce the geometric proof. The analytic solution is straight-forward, while the geometric solution required experimentation and insight. It may not be as beautiful, but the analytic solution is definitely simpler.

championmathgirl · 2015-07-23 08:34:33

Those are great proofs. But is there a way using coordinate geometry? For example putting triangle ABC in the coordinate plane so that A is the origin and one of the equal sides aligns with the x-axis.

I tried to go on from there, but I'm not sure what to do.

Bob · 2015-07-23 19:42:38

Cannot spare the time now but, if no one else manages this, I'll have a try next week.

Bob

Bob · 2015-07-27 02:38:13

Tied up in horrible algebra at the moment. In theory it can be done this way but I need to find some errors in my working first.

Bob

Bob · 2015-07-27 06:19:42

OK. Here's the coordinate geometry method:

strategy: simplify whenever possible to keep the algebra simple (relatively) and work through the given information, getting all the coordinates and using product of gradients of perpendicular lines = -1.

Let A be (0 , 0) and B (b , 0). Also C (c , d). As AB = AC =>

D is the midpoint of BC so coordinates are

gradient of AC = d/c => gradient of ED = -c/d

=> equation of ED is

equation of AC is y = dx/c so equating the y values to get the x coordinate of E

So, gradient of BE is

and gradient of AF is

The product of these gradients is

So AF is perpendicular to BE.

Bob

Last edited by Bob (2015-07-27 06:41:38)

rileywkong · 2016-07-20 06:02:13

For some reason Geogebra disagrees with the fact that AF is perpendicular to BE...

Last edited by rileywkong (2016-07-20 06:02:39)

bobbym · 2016-07-20 06:44:30

Hi rileywkong;

I am using it and it verifies that the angle is 90 degrees.

thickhead · 2016-07-20 19:54:33

If you want to use coordinate geometry it is simpler if D is taken as (0,0) A as (0,a), C as (c,0) and B as(-c,0)

Last edited by thickhead (2016-07-20 19:59:05)

dazzle1230 · 2016-07-21 08:56:53

this is what I did so far:
Let the coordinates of the points in triangle ABC be A(0,0), B(a,0), and C(c,d). Since D is the midpoint of BC, the coordinate points of D are

. We then know that slope of AC is . Since DE is perpendicular to AC, we know that the slope of DE is .

I was planning to make equations for lines AC and DE and find the intersection which would be the coordinates for point E. I made an equation for line AC which is y=d/c x. However, when I tried to create an equation for line DE, the y-intercept seemed very complicated: y=-c/d x+(ac+c^2+d^2)/2d. I tried solving for x and y, but the fractions didn't come out as easy numbers to work with.

Can someone see if I made an error?

Last edited by dazzle1230 (2016-07-21 08:58:32)

bobbym · 2016-07-21 15:18:54

Hi;

For one thing you can be somewhat tighter when defining C.

thickhead · 2016-07-21 15:54:40

Hi dazzle 1230,
while selecting coordinate system you have to select the one which makes your task simple. sine the triangle is isosceles either AD must be along y axis or x axis. Origin must be either A or D. I found D more convenient since I had to deal with the equation of DE. Also bobbym has a point shown above.

thickhead · 2016-07-21 17:42:15

Hi dazzle 1230,
Why don't you take C on x axis? DE will be vertical.

dazzle1230 · 2016-07-22 01:21:31

Thanks for all the help! It made the problem so much easier!

dazzle1230 · 2016-07-22 01:53:17

One more question:
as my final result, the slopes of AF and BE are

and respectively. Are those accurate because they don't look like negative reciprocals...

bobbym · 2016-07-22 02:52:14

Check those slopes again.

thickhead · 2016-07-22 04:09:48

hi dazzle 1230,

dazzle1230 · 2016-07-22 05:09:23

I don't understand how you got a^2-c^2=-b^2

Math Is Fun Forum

#1 2014-03-14 16:19:12

Triangles

#2 2014-03-14 23:29:48

Re: Triangles

#3 2014-03-15 19:55:41

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#4 2014-03-15 19:57:48

Re: Triangles

#5 2014-03-17 14:24:50

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#6 2014-03-18 03:14:39

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#7 2014-03-19 20:51:39

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#8 2014-03-21 23:52:37

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#9 2014-03-23 06:52:18

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#10 2015-07-23 08:34:33

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#11 2015-07-23 19:42:38

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#12 2015-07-27 02:38:13

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#13 2015-07-27 06:19:42

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#14 2016-07-20 06:02:13

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#15 2016-07-20 06:44:30

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#16 2016-07-20 19:54:33

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#17 2016-07-21 08:56:53

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#18 2016-07-21 15:18:54

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#19 2016-07-21 15:54:40

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