As shown in the figure, ABC = ADC = 90 °, M is the midpoint of AC, Mn ⊥ BD and N, and BN = nd is proved ditto

prove:
∵∠ABC=∠ADC=90°
The ⊿ ABC and ⊿ ADC are right triangles, and AC is a common hypotenuse,
∵ m is the midpoint of AC
The BM and DM were the median lines of hypotenuse, respectively
Ψ BM = DM = ½ AC [right triangle, the middle line of hypotenuse equals half of hypotenuse]
The MBD is an isosceles triangle
∵ Mn ⊥ BD according to the three lines in one, Mn is the bottom midline
∴BN=ND

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