order to show how the angles relate to one another, we will need to use a
theorem about a different case from what we see here. Consider the case below.
claim that <ACB is 2<ADB.
What I am saying is this: When we have two triangles that share the same
chord of a circle, both triangles on the same side of that chord, and one
triangle having the center of the circle as the third point and the other
having a point on the circle, then no matter where the point is on the circle,
the angle will never change.
this is obviously a big claim that needs to be proved, so letŐs do this. LetŐs connect points C and D.
construction, we have three different isosceles triangles. LetŐs call <BDC = <DBC = x,
<CDA = <DAC = y. Then
<BCD = 180 – 2x, <ACD = 180 – 2y. We know that the total degrees around C add up to 360,
therefore <ACB = 360 – (180 – 2x) – (180 – 2y) which
simplifies to 2x + 2y. This is
exactly two times <ADB (which was x + y). This proof works for cases that look like this (where the
two triangles do not intersect), so letŐs look at the other case.
again, I have connected C and D.
We still need to show that <ACB is 2<ADB. We have three isosceles triangles (ACD,
ACB and DCB) so letŐs label some of these angles. Let <BDC = <DBC = x. Let <DBA = y so that <BAC = x + y. Then <ACB = 180 – 2x –
2y. Also, let <BDA = z so that
<DAC = x + z. Then from
triangle DAB, we can see that z + (x + z) + (x + y) + y = 2x + 2y + 2z = 180. Therefore our <ACB = 2z which is two
times <ADB, which is what we wanted to prove.
LetŐs go back and take a look at our original problem now and use this
is not apparent from this picture, but let me dissect a part for you to show you
what I am going to do. Take a look
Without moving the points, I have erased all the lines that we are not
going to use. I have also added in
a segment (AD) to let you see how the theorem is going to be used. Triangles ACD and AFD both share a
chord AD. From above, I can state
that since C and F are on the circle on the same side that the angles are
congruent. Using this idea
repeatedly, we can see that the correlation goes as follows.
we allow <A = 2x, <B = 2y and <C = 2z (since we constructed the other
triangle from angle bisectors), then <D = x + y, <E = y + z, and < F =
x + z.
nice thing about using this proof is that as long as the requirements are met
for the proof, then it doesnŐt matter what kind of triangle you begin with.