Archimedes problem about the broken chord

Question

In a circle of radius 5 is inscribed triangle ABC and AB = 4, BC = 2. From the middle of the lesser of the two arcs AC (point K) lowered perpendicular to the chord AB. Prove that AM = MB + BC.

Actually it's a quite simple problem, but I faced the difficulty in another thing. This is a right drawing, where required feature (AM = MB + BC) is satisfied (angle ABC is obtuse):

But there's another one drawing which fully satisfies with condition, however it's an obvious that AM < MB + BC (angle ABC is acute):

How to prove that in the second case required feature isn't satisfied?

Answer 1

Here's the correct image :) (that big circle arc is the large circle)

Okay now onto the question. This is something known as 'diagram dependency' - dependency on your version of the diagram. Now the proof for AM' = M'B + BC' obviously fails because C lies between A and B so M' is closer to A than B.

EDIT: After seeing @rschwieb 's answer note that $M' = M$ and $BC = BC'$ so both versions are clearly the same. Just prove that $M' = M$

Answer 2

Comparing your statement with the top hits of a Google search reveals a discrepancy. Everything I found says "from the middle of the arc containing the angle $ABC$" rather than "the lesser of the two arcs." That seems to be the problem with the second drawing.

(There is then some stipulation about dropping the perpendicular on the larger of the two chords.) So possibly your source just has a mistaken statement.

Answer 3

Well, in the second case, when $K$ is on the arc different from the arc $ABC$ $$AM = MB - BC$$ so definitely $AM = MB - BC < MB + BC$

And by the way, this is a very general problem, these values $5, 4, 2$ play no role in the proof.

Let $D$ be chosen on the line $AB$ so that $BD = BC$ and $B$ is inside the segment $AD$. Then triangle $BCD$ is isosceles. Since $K$ is the midpoint of the arc $ABC$, the line $BK$ is the exterior angle bisector of angle $\angle \, ABC$ and is thus the (interior) angle bisector of angle $\angle \, CBD$. As an angle bisector of the isosceles triangle $BCD$, the line $BK$ is also the orthogonal bisector of edge $CD$. Therefore, $KC = KD$. As mentioned before, $K$ is the midpoint of arc $ABC$ so $KA = KC$. Hence, $KA = KC = KD$ and thus triangle $ADK$ is isosceles. Then $KM$ is its altitude, so it is also the orthogonal bisector of $AD$. Hence, $AM = MD$. However, by construction $BC = BD$ so $$AM = MD = MB + BD = MB + BC$$

Now let us take a look at point $K'$ and its orthogonal projection $M'$ on $AB$. Analogously to the previous case, let $D'$ be chosen on the line $AB$ so that $BD' = BC$ and $D'$ is inside the segment $AB$. Then triangle $BCD'$ is isosceles. Since $K'$ is the midpoint of the arc $AC$, not containing $B$, the line $BK'$ is the angle bisector of angle $\angle \, ABC$. As an angle bisector of the isosceles triangle $BCD'$, the line $BK'$ is also the orthogonal bisector of edge $CD'$. Therefore, $K'C = K'D'$. As mentioned before, $K'$ is the midpoint of arc $AC$, so $K'A = K'C$. Hence, $K'A = K'C = K'D'$ and thus triangle $AD'K'$ is isosceles. Then $K'M'$ is its altitude, so it is also the orthogonal bisector of $AD'$. Hence, $AM' = M'D'$. However, by construction $BC = BD'$ so $$AM' = M'D' = M'B - BD' = M'B - BC$$