Question

In an isosceles triangle ABC with AC = BC, point M is on the side BC, and point N is on the segment MC. It is known that MN=AM, and ∠BAM=∠NAC. Find m∠MAC, in degrees.

Answer 1

m∠MAC is equal to m∠NAC.

Step-by-step explanation:

Let's solve the problem step by step:

1. Given that ∠BAM = ∠NAC and AC = BC, we can conclude that triangle ABC is an isosceles triangle.
2. Since AC = BC, we also have AM = BM (isosceles triangle property).
3. Now let's consider triangle MCA. Since MN = AM, we can deduce that triangle MNA is also an isosceles triangle.
4. As MNA is an isosceles triangle and MN = AM, we can conclude that ∠MAN = ∠MNA. Since ∠BAM = ∠NAC, it implies that ∠MAN = ∠NAC.
5. By combining both results, we have ∠MAN = ∠MNA = ∠NAC.
6. Since AC = BM and ∠MAN = ∠NAC, we can conclude that triangle ACM is congruent to triangle BMA (by SAS congruence).
7. As a consequence of triangle congruence, the corresponding angles are congruent. Hence, ∠MAC = ∠MAB.
8. Since triangle ACM is congruent to triangle BMA, we have ∠MAC = ∠MAB = ∠BAM.
9. Finally, since ∠BAM = ∠NAC, we can deduce that ∠MAC = ∠MAB = ∠BAM = ∠NAC.

Therefore, m∠MAC is equal to m∠NAC.