In the math interest group activities, the students explored the folding problem of diamonds. Among them, an interesting question is: in the diamond-shaped ABCD, ∠ ABC = 70, e is the midpoint of BC, ∠ EAB = 40, and find the degree of ∠AEB.
First of all, we can get AB = BC∠ABC = 70 according to the nature of diamonds, so ∠ cab = 50. Because e is the midpoint of BC, AE⊥BC can be obtained according to the midline theorem of isosceles triangle. Meanwhile, since ∠ EAB = 40, ∠ BAE = 180-90-40 = 50.
Next, according to the triangle interior angle sum theorem, we can get ∠ AEB = 180-50-50 = 80. And because AE divides ∠BAC equally, we can get ∠ BAE = ∠ EAC = 40. Therefore, we can get ∠ EAB = 180-40-80 = 60.
Another method is to use the properties of diamond and congruent triangles to solve it. In the diamond ABCD, AB=BC, so ∠ACB=∠BAC. And because AE divides ∠BAC equally, we can get ∠ BAE = ∠ EAC = 40. Therefore, we can get △ Abe△ ace, that is, BE=EC.
Because e is the midpoint of BC, BE=EC=BC/2. And because of △ Abe △ ace, we can get AB=BE=EC=BC/2. Therefore, we can get that △ABE is a right triangle and △ ∠BAE is a right angle. Therefore, we can get ∠ AEB = 180-40-90 = 50.
Through the above two methods, we can get that the degree of ∠AEB is 50 or 60 or 80. It should be noted that the results of these two methods are not contradictory, because there may be different solutions under different conditions. Therefore, in practical application, it is necessary to choose appropriate methods to solve specific problems.