Firstly, the length of AC is calculated by Pythagorean theorem, and then the length of CD is calculated. Then the median line on the hypotenuse of the right triangle is equal to half of the hypotenuse, and the answer is obtained. The second problem is that △ADE is an equilateral triangle, and then it is found that point E, D' is symmetrical about the straight line AC, and the connecting line DD' intersects AC at point P, where the value of DP+EP is the smallest, and then the answer is obtained;
Solution: (1) ∵∠ BAC = 45, ∠ B = 90,
∴AB=BC=6 pieces, 2cm, ∴AC= 12cm,
∫∠ACD = 30,∠DAC=90,AC= 12cm,
This is the detailed answer/exercise/math /8006 17. Put a pair of right triangles together to get a quadrilateral ABCD, where ∠ BAC = 45, ∠ ACD = 30, point E is the midpoint of CD edge, connect AE, and fold △ADE along the straight line where AD is located to get △ ad' e, D. If AB=6, the root number is 2cm.
The length of AE is (2) Try to determine a point P on the line segment AC to minimize the value of DP+EP.
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