Good day,
I have a question regarding geometry. I don't know whether my answer is correct because the answer in my book uses a totally different method for solving this particular problem.
Here's the problem:
Given is a triangle, ABC, in which the middle of AB is also the middle of the circle drawn outside of the triangle. Finally, we draw a line from C to M where M signifies the point in the middle of AB. Now you have angle C sub 1 and angle C sub 2 ( together: angle C sub 12.)
We have created 2 equal-sided triangles, within the triangle ABC, within the circle, namely: triangle ACM and triangle BCM.
Proof that angle C sub 12 is equal to 90 degrees.
My answer:
Given:
AM = MC = MB (2 equal-sided triangles)
To Proof:
$$\angle C_{12} = 90^{\circ}$$
Proof: $$ \left.\begin{matrix} & & & & & \\ & & & & & \\ & & & & & \\ \angle A + \angle M_{1} + \angle C_{1} = 180^{\circ}\\ \angle B + \angle M_{2} + \angle C_{2} = 180^{\circ}\\ \angle M_{2} + \angle 2B = 180^{\circ} \\ \angle M_{1} + \angle 2A = 180^{\circ} \\ \angle M_{1} + \angle M_{2} = 180^{\circ}\\ \angle A + \angle B + \angle C = 180^{\circ} (= Q)\\ \\ \end{matrix}\right\}general \\ $$
Calculate for angle A and angle B:
$$ \angle M_{2} + \angle 2B = 180^{\circ}\\ \angle 2B = 180^{\circ} - \angle M_{2}\\ \angle B = 90^{\circ} - (\angle M_{2} / 2) $$
$$and$$
$$ \angle M_{1} + \angle 2A = 180^{\circ}\\ \angle 2A = 180^{\circ} - \angle M_{1}\\ \angle A = 90^{\circ} - (\angle M_{1} / 2) $$ Also: $$ \angle M_{2} = 180^{\circ} - \angle M_{1}\\ \angle M_{1} = 180^{\circ} - \angle M_{2} (= U)\\ $$ Now we can put angle A and angle B in the sum of angles (indicated by Q in general section.):
$$(90^{\circ} - (\angle M_{1} / 2)) + (90^{\circ} - (\angle M_{2} / 2)) + \angle C = 180^{\circ}$$
Now we substitute $$ \angle M_{1}$$ inside $$((\angle M_{1} / 2)) $$ with U:
$$(90^{\circ} - (\frac{180^{\circ} - \angle M_{2}}{2}) + ((90^{\circ} - (\angle M_{2} / 2)) + \angle C = 180^{\circ}\\ (90^{\circ} - (\frac{180^{\circ}}{2}-\frac{\angle M_{2}}{2})) + (90^{\circ} - (\angle M_{2} / 2)) + \angle C = 180^{\circ}\\ 90^{\circ} - \frac{180^{\circ}}{2}+\frac{\angle M_{2}}{2} + 90^{\circ} -\frac{\angle M_{2}}{2} + \angle C = 180^{\circ}\\ \Rightarrow 90^{\circ} + \angle C = 180^{\circ}\\ Conclusion: \angle C_{12} = 90^{\circ} $$
(not the prettiest proof but hopefully you get the idea.)
Here's the more concise ( and beautiful ) proof from the answers in my book:
$$\angle A + \angle B + \angle C = 180^{\circ} \Rightarrow \angle C_{1} + \angle C_{2} + \angle C_{12} = 180^{\circ} \Rightarrow \angle 2C_{12} = 180^{\circ} \Rightarrow \angle C_{12} = 90^{\circ} $$
If you need more information please ask and I'll provide the necessities.
Thanks!
Picture:
Note: My figure is misleading. ABC in figure is not isosceles.
$$2x+2y = 180^\circ$$
$$x+y = ?$$