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Solution of the finale problem of mathematical geometry in the second day of junior high school
(1)AB=4, C(2 radical numbers 3+2, 2 radical numbers 3);

(2)AE=t OE=2-t,BD= 2t,AD = 4-2t;

If AEFD is a diamond,

t=4-2t,

t=4/3

(3) If PA=PC, then P(2 3,0);

If AP=AC, then P(2 7,0) P (-2 7,0);

If PC=AC, then P(2 3+2-2 5,0).