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## #1 2014-07-22 14:28:35

harrychess
Member
Registered: 2014-04-04
Posts: 33

### Medians and Altitudes

Help please! An answer and explanation will do.

In quadrilateral ABCD, AB=BC=13, CD=DA=24, and angle D=60.  Points X and Y are the midpoints of BC and DA respectively.  Compute XY^2 (the square of the length of X).

Last edited by harrychess (2014-07-22 14:52:24)

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## #2 2014-07-25 09:59:40

bob bundy
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Registered: 2010-06-20
Posts: 8,322

### Re: Medians and Altitudes

hi harrychess,

Hmm.  Working on it.

So far:

BD is a line of symmetry for the kite.  AC is perpendicular to BD and equal to 24.  ADB = 30.

In my diagram if Z is at the intersection of XY and BD, it looks like XZ = ZY, but I cannot see why just yet.

Hopefully more will follow when I've caught up from jet lag.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #3 2014-07-26 00:38:55

bob bundy
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Registered: 2010-06-20
Posts: 8,322

### Re: Medians and Altitudes

I think this will do it.

The shape is a kite and BD is the line of symmetry.  AC is at right angles to BD.

Z is the point where BD cuts XY.  I have also reflected XY in the line BD to create points X' and Y'.

Because of the symmetry YY' = XX' = 12, and both these lines are perpendicular to BD **.  Thus X'YY'X is a rectangle and so XZ = XY.

Thus you can get XY by first using Pythag on triangle YY'Z.

Bob

** edit. Certainly YY' = 12 by equilateral triangles.  I'm thinking I may be assuming I have 'proved' XX' is also 12 when I may not yet have done so.  Just improving my proof now.

Last edited by bob bundy (2014-07-26 00:47:50)

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #4 2014-07-26 00:47:49

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037

### Re: Medians and Altitudes

Hi Bob

How would you use Pythag on YY'Z? It's not a right triangle.

Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

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## #5 2014-07-26 01:22:54

bob bundy
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Registered: 2010-06-20
Posts: 8,322

### Re: Medians and Altitudes

hi Stefy,

Yes, you're right.  The above is flawed.  I'm working on a corrected version.  At the moment X'YY'X is an isosceles trapezium.  I'm sure I had it sorted a while back but I cannot re-trace my steps.  I had a way of showing XZ = ZY and that forces the trapezium to become a rectangle.  Then you probably have to use cosine rule.

Also DYC = 90.  This should help but I need angle YCX.

Thinking still ........................................

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #6 2014-07-26 01:29:07

bob bundy
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Registered: 2010-06-20
Posts: 8,322

### Re: Medians and Altitudes

OK.  Like this:

AC = 24.  Use cosine rule to calculate ABC and hence BCD.  Subtract 30 and you have YCB.  Then use cosine rule on XYC.

Checking ..........................

Bob

Last edited by bob bundy (2014-07-26 01:30:55)

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #7 2014-07-26 01:49:08

bob bundy
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Registered: 2010-06-20
Posts: 8,322

### Re: Medians and Altitudes

A better version.

Because YD = 12 and CD = 24, that makes CYD a 30-60-90 triangle.

Angle ACB can be found thus:

cosine ACB = 12/13 and thus sine ACB = 5/13.

cosine BCD = cosine(BCA + 60) = cosBCA.cos60 - sinBCAsin60 = (12/13 - 5root3/13)/2 = (12 - 5root3)/26

***find sine BCD.  Running out of time now, I'll fill this in later.

cosine XCY = cosine(BCD - 30) = cos BCD.cos30 + sinBCD.sin30

Then use cosine rule on XCY.

Note:  I've used compound angle formulas to avoid the 'inaccuracy' of using a calculator for these angles.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #8 2014-07-26 03:00:12

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037

### Re: Medians and Altitudes

Haven't checked your new solution yet, but XX'YY' is not an isosceles trapezium, it's a parallelogram.

EDIT: Actually, yes, it's both, but calling it a parallelogram is more restrictive.

EDIT #2: And, yes, it is also a rectangle

Last edited by anonimnystefy (2014-07-26 03:07:34)

Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

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## #9 2014-07-26 05:37:16

bob bundy
Administrator
Registered: 2010-06-20
Posts: 8,322

### Re: Medians and Altitudes

It is my hope to prove it is a rectangle.  So far I have only proved the isosceles trapezium.  I am about to tidy up and finish off my method.  It will take a little while.

What this space.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #10 2014-07-26 06:02:59

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037

### Re: Medians and Altitudes

Well, the opposite sides are equal and parallel...

Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

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## #11 2014-07-26 07:43:32

bob bundy
Administrator
Registered: 2010-06-20
Posts: 8,322

### Re: Medians and Altitudes

Stefy wrote:

Well, the opposite sides are equal and parallel...

From my diagram and analysis YY' is parallel to XX' (as both are perpendicular to BD) and X'Y = XY' (by reflection).  I don't yet have evidence that X'Y is parallel to XY' nor that X'X = YY'.

But I have simplified my calculations:

Let BCA = a and XCY = b

By simple trig cos(a)=12/13 and sin(a) = 5/13

Therefore, cos(b) = cos(a+30) = cos(a)cos(30) - sin(a)sin(30)= (12root3 -5)/26

Then XY^2 = 144x3 +169/4 -6root3(12root3-5) = 310.2115

This is confirmed by my Sketchpad drawing.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #12 2014-07-26 08:27:40

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037

### Re: Medians and Altitudes

All of the edges of XX'YY' are midsegments of respective triangles made by diagonals of the larger quadrilateral.

Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

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## #13 2014-07-27 06:38:28

bob bundy
Administrator
Registered: 2010-06-20
Posts: 8,322

### Re: Medians and Altitudes

Arhhh.  You say that like it is a theorem I should know.  I don't recall ever meeting that one, but that's not surprising as I don't bother to learn theorems.  But it was easy to prove, so I'll add it to the list of theorems that I won't bother to learn.

Then it quickly follows that the shape is a rectangle.  I expect you can get XY^2 using that, but I don't think I'll bother now.  I'll leave it to the interested reader.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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