As shown in the picture (the painting is not good, let's have a look). Isosceles triangle ABC, base BC= 15cm, and CD is the center line of waist AB. In both cases (1), the circumference of triangle ACD is larger than that of triangle BCD. Have AC+AD+CD-(BD+CD+BC) = AC+AD+CD-BD-CD? Bring BC= 15? AD=BD can draw a conclusion? AC- 15=8？ AC=23 In the same way, it can be concluded that in the second case, the perimeter of triangle BCD is larger than that of triangle ACD, and AC=7 can also be obtained. But according to the sum of two sides of the triangle is greater than the third side 7+7 < 15, BC=23?

The second question is 2a?

Or does this picture regard CD as a high line, ∠B=∠ACB? So ∠BCD=a? ∠A= 180 degrees -∠B-∠ACB= 180 degrees -(90 degrees -∠B)-∠C? =? 180 degrees -(90 degrees -a)-? ∠B= 180 degrees -(90 degrees -a)? -(90 degrees -α)= 2a

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