my solution:anemone said:
Hello.Albert said:
In case there are other people like me who took a while to follow Albert's clever proof, here is how it works. You are given the angles of triangle ABC: $103^\circ$ at A, $51^\circ$ at B and therefore $26^\circ$ at C. You are also given the angles of triangle AMC: $30^\circ$ at A, $13^\circ$ at C and therefore $137^\circ$ at M. Since 13 is half of 26 it follows that MC bisects the angle at C. So if we drop perpendiculars E, F from M to AC and BC, the right-angled triangles MEC, MFC will be congruent, with angle $77^\circ$ at M. Now draw the circle centred at M touching AC, BC at E and F, and let BG be the other tangent to the circle from B. Then the right-angled triangles BFM, BGM are congruent. At this stage we know the angles $60^\circ$, $77^\circ$, $77^\circ$ at M. So the remaining angle AMF is $146^\circ$ and this is bisected by BM. Therefore the angle BMF is $73^\circ$; and its complement, the angle MBC, is $17^\circ.$[/sp]Albert said:
That is a very good point, which I had completely overlooked. The only way round it that I can see is to do essentially as you suggest. Namely, suppose that the tangent at G meets the line BC at some point B', and then show that B'=B.johng said:
That implies that you are defining G as the point where the tangent from B meets the circle. In that case, you need to explain why G lies on the line AM.Albert said:
Yes, that seems to work. You are saying that the angles BAG and BAM are both $73^\circ$. So G (defined as the point where the tangent from B meets the circle) lies on the line AM.Albert said:
Albert said:
Yes, you are right again, and this time I don't see any way to fix it. The distorted diagram showing G and G' as distinct points highlights the difficulty in Albert's argument. Fortunately, anemone has now given us a watertight proof using Ceva's theorem.johng said:
Albert said:
please see post #9anemone said:
Post #9 states that $\angle FBG=\dfrac{77+77+60-146}{2}$. I believe that is the step that several of us find incomprehensible. Could you explain it please?Albert said:
View attachment 1839Opalg said:
The measure of angle MBC in Triangle ABC can be found using the Law of Cosines or the Law of Sines. The Law of Cosines states that c^2 = a^2 + b^2 - 2abcosC, where c is the side opposite angle C. The Law of Sines states that sinA/a = sinB/b = sinC/c. By plugging in the known values for sides and angles, the measure of angle MBC can be calculated.
The proof for finding the measure of angle MBC in Triangle ABC involves using the properties of triangles, such as the Triangle Sum Theorem and the Exterior Angle Theorem. By constructing a line parallel to side BC through point A, a triangle with known angles can be formed, allowing for the measure of angle MBC to be calculated using the Triangle Sum Theorem and the Exterior Angle Theorem.
The steps for finding the measure of angle MBC in Triangle ABC involve first identifying the known sides and angles, then using the Law of Cosines or Law of Sines to calculate the measure of angle MBC. Alternatively, the proof method can be used by constructing a line parallel to side BC through point A, and then using the properties of triangles to calculate the measure of angle MBC.
Yes, the measure of angle MBC can also be found using the Triangle Sum Theorem and the Exterior Angle Theorem. By constructing a line parallel to side BC through point A, a triangle with known angles can be formed, allowing for the measure of angle MBC to be calculated using these theorems.
Yes, there are special cases where the measure of angle MBC cannot be determined. One example is when Triangle ABC is a right triangle, as angle MBC would be the right angle and its measure would be 90 degrees. Another example is when there is not enough information given about the sides and angles of Triangle ABC to use the Law of Cosines, Law of Sines, or the proof method.
