ABCD is a quadrilateral that is incribed in a circle and that has an
incircle. EF is a diameter of the circle with EF perpendicular to BD.
BD intersects EF at M and AC at S. A and E lie on the same side of BD.
Prove that AS:SC = EM:MF.

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Clarification of Question by g7steve-ga on 05 Nov 2004 19:06 PST

Mathtalk - please post your answer as the official answer.  Thanks. 
BTW, this question is coming from my old high school friend(as high
school teacher now).

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Request for Question Clarification by mathtalk-ga on 06 Nov 2004 04:57 PST

Thanks, g7steve-ga, but please clarify whether EF is a diameter of the
outer circle (in which ABCD is inscribed) or a diameter of the
"incircle" of ABCD.

regards, mathtalk-ga

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Clarification of Question by g7steve-ga on 06 Nov 2004 09:46 PST

Yes, EF is a diameter of the outer circle.

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Clarification of Question by g7steve-ga on 09 Nov 2004 05:52 PST

I am not sure it helps or not:

you don't have to prove that it's a kite!
 
If there is a thoery that diagonals in a quadrilateral with incircle
are perpendicular to each other, then I could do the rest.
 
Reason:
 Because BD perpen. to EF
and, iff, AC perpen. to BD,
thn AC parallel to EF hence by constructing extra trianges,
AS:BS = EM:BM and
CS:BS = FM:BM 
HENCE AS:SC = EM:MF

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Request for Question Clarification by mathtalk-ga on 09 Nov 2004 06:20 PST

Ah, this is very helpful.  I was laboring under the mistaken
impression that S was where EF intersected AC (rather than where BD
intersects AC), but every picture I sketched came up with EF and AC
looking very parallel (just as your conjecture predicts).

I think it is not true that the diagonals AC and BD are perpendicular.
 I picked some odd angles a = 0.3 and b = 0.8 as described in the
Comment below, and checking the dot product between the diagonal
direction gives a nonzero result (so they aren't perpendicular).

regards, mathtalk-ga

If you can prove that a quadrilateral that is inscribed in a circle
and has an incircle is symmetrical about one of its diagonals (that
is, it is a kite shape) the rest is easy.  I don't know how to prove
the symmetry, but having an incircle means the angle bisectors of all
four vertices intersect at a single point, and having an 'outcircle'
means that the perpendicular bisectors of all four sides intersect at
a single point.

Where a general quadrilateral has five degrees of freedom (e.g. four
sides plus one angle), a quadrilateral which is inscribed in a circle
has only four degrees of freedom, and a quadrilateral that is both
inscribed in a circle and cirumscribed on a circle (what I assume it
means to have an incircle) will have three degrees of freedom (e.g.
three sides or equiv. two angles and an included side).

A quadrilateral is inscribed in a circle iff opposite angles are
supplementary.  (If this is true of one pair of opposing interior
angles, it is also true of the other pair.)

Therefore the only such quadrilaterals which are symmetric about a
diagonal are formed by the reflection of a right triangle on its
hypotenuse.  This gives us only two degrees of freedom, so it doesn't
exhaust all possibilities.

regards, mathtalk-ga

That is true, I apologize.  The kite hypothesis is false.  By
inscribing a rectangle, taller than it is wide, in a circle, and then
shifting the bottom of the rectangle upward, keeping the vertices on
the circle, you will eventually (when the combined lengths of the
sides equals the combined length of the top and bottom) reach a shape
which has an incircle.  The shape is a trapezoid and cannot be a kite.

Not an easy problem.  Where did it come from?

Here's a fairly easy way to parameterize all the quadrilaterals that
can both be inscribed in a circle and have an incircle (be
circumscribed around a circle).

We need at least one "length" parameter for scale of the figure.  I
will take the radius r of the "incircle" as this parameter (one might
fix this radius as 1, then scale the resulting figure to whatever size
is desired).

The other parameters are two acute angles a and b.  (I could post
Greek alpha and beta here, because Google Answers supports Unicode
characters, but it might prove inconvenient for some Readers.)

The general quadrilateral of the type considered here can then be
decomposed into four "kite" shapes of racecar-ga's description, or
eight right triangles, as follows:

Let O be the center of the incircle, and let the acute angle a and b
appear at the respective vertices A and B of triangle OAB.  Drop a
perpendicular from O to side AB. This will form a leg of length r for
both right triangles having opposing angles a and b respectively.

Reflect each of these two triangles across their hypotenuse, forming
two "kite shapes" joined at O along the original radius of length r. 
This creates two new radii from O, and the hypotenuses leading from O
are angle bisectors at A and B respectively.

Extend the new leg from A beyond the radial to a point D where a right
triangle is formed with hypotenuse OD, and the angle at D (opposite
the radial leg perpendicular to AD) is comlementary to b, say b'. 
Similarly extend from B to a new vertex C along that kite's outer
edge, forming a right triangle with hypotenuse OC and angle at C
complementary to a, say a'.

Now the quadrilateral ABCD is defined by these four points, and its
decompostion into eight right triangles is completed by reflecting the
final two right triangles constructed across their hypotenuses, OD and
OC.  The final two radial legs coincide, as may be seen by addition of
the various angles around O.

For from the original two right triangles (with distil angles a and b)
we have central angles a' and b', and doubling these, then counting
the contributions from the final four four triangles:

(a' + b') + (a' + b') + (a + b) + (a + b)

        = 2 (a + a' + b + b')

        = 2 (pi/2 + pi/2) = 2pi

give one complete revolution.

That ABCD can be inscribed in a circle follows from the angle 2a at A
being supplementary to the angle 2a' at C, or equally that angle 2b at
B is supplementary to the angle 2b' at D.

regards, mathtalk-ga

To clarify a point raised by g7steve-ga above, the diagonals (of such
quadrilaterals as we construct above) are perpendicular if and only if
a and b are complementary.  This is not generally true, as a,b could
be any acute angles.

regards, mathtalk-ga

Sorry, my statement above is incorrect.  I've confused the diagonal AC
and BD with the triangulation I constructed of the quadrilateral. 
Ooops!

--mathtalk-ga

Here are the expressions for the vertices of the quadrilateral in
terms of acute angles a and b, as described above.  We assume the
"incircle" is centered at the origin and has radius 1, taking side AB
perpendicular to the positive x-axis:

A = [ 1, cot(a) ]

B = [ 1, -cot(b) ]

C = sec(a) * [ -cos(a-2b), sin(a-2b) ]

D = sec(b) * [ -cos(b-2a), -sin(b-2a) ]

With this "synthetic geometry" approach it should be possible to
decide the proposition in a brute force way.  Perhaps it will also
give insight into a proof from Euclid's axioms.

For example, point M is simply (B + D)/2 and once the center O of the
"out-circle" is found, its diameter EF passing through M is
determined.

regards, mathtalk-ga

The first time I saw your problem here, last 08Nov04, I got as far as
trying to prove that secant AC is parallel to diameter EF. My three
sketches showed that the two seem to be parallel. It is the AS:SC =
EM:MF that I just cannot hurdle.

Now in one of your Clarifications, you said if the two were parallel,
and so AC is perpendicular also to BD, then
AS:BS = EM:BM
How can that be? Right triangles EMB and ASB are not similar. 

CS:BS = FM:BM
Again, cannot be. Right triangles BMF and BSC are not similar also.

To show more my point, let us get another right triangle on the same
side of BD as right triangles EMB and ASB. Let it be closer to B. Let
us call it triangle UVB where BV is on the secant BD and U is on the
circumcircle. Here UV is a lot longer than BV. Following your
reasoning above, AS:BS = EM:BM = UV:BV? No.

This is not to question your know-how. This is just to show how
difficult it is to prove that AS:SC = EM:MF.
They look proportional, allright, but to prove....

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Or, I just don't know how you arrived at AS:BS = EM:BM and CS:BS = FM:BM.
Please let me know.