In the triangle ABC, D is the midpoint of AB, AC = 12, BC = 5, CD = 13 / 2 As the question, fast, 1 hour to answer Don't Scribble Please look at the title and the process carefully. If there is any mistake in the answer, please answer the question carefully, My idea is to draw another right triangle to prove their congruence

In the triangle ABC, D is the midpoint of AB, AC = 12, BC = 5, CD = 13 / 2 As the question, fast, 1 hour to answer Don't Scribble Please look at the title and the process carefully. If there is any mistake in the answer, please answer the question carefully, My idea is to draw another right triangle to prove their congruence

I don't know if you've studied cosine theorem
According to the cosine theorem:
AB^2+BC^2-2*BC*AB*∠B=AC^2
BC^2+BD^2-2*BC*BD*∠B=CD^2
Then the data is substituted into:
AB^2+25-10*AB*∠B=144 ①
(1/2*AB)^2+25-5AB*∠B=169/4 ②
①-②*2 1/2*AB^2-25=144-169/2
It is reduced to ab ^ 2 = 169
According to ab ^ 2 = BC ^ 2 + AC ^ 2, it can be concluded that △ ABC is a right triangle
Take the midpoint of AC as e and connect De
According to the median line parallel to the bottom and equal to half of the bottom: de = 1 / 2 * BC = 5 / 2
According to Pythagorean theorem, it can be obtained that △ CDE is a right triangle and ∠ CED = 90 degrees
According to the parallel line theorem: two straight lines are parallel, and the same angle is equal. ∠ ACB = 90 degrees
So △ ABC is a right triangle