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[1] | 1 | ;;
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[99] | 2 | ;; Prove Apollonius Circle Theorem:
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| 3 | ;;----------------------------------------------------------------
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| 4 | ;; If ABC is a right triangle with hypotenuse BC,
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| 5 | ;; and
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[1] | 6 | ;;
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[99] | 7 | ;; 1) M is the midpoint of BC;
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| 8 | ;; 2) M1 is the midpoint of AB;
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| 9 | ;; 3) M2 is the midpoint of AC;
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| 10 | ;; 4) is the foot of the altitude dropped from A;
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| 11 | ;;
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| 12 | ;; then A, H, M1, M2 and M lie on the same circle.
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| 13 | ;;----------------------------------------------------------------
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| 14 | ;;
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| 15 |
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[1] | 16 | (prove-theorem
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[99] | 17 |
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| 18 | ;; If
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| 19 | (
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[101] | 20 | (perpendicular A B A C) ; AB _|_ AC
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| 21 | (midpoint B C M) ; M is the midpoint of BC
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| 22 | (midpoint A M O) ; O is the midpoint of AM
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| 23 | (collinear B H C) ; H lies on BC
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| 24 | (perpendicular A H B C) ; AH _|_ BC
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[99] | 25 | )
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| 26 | ;; Then
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| 27 | (
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[101] | 28 | (equidistant M O H O) ; MO = HO
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| 29 | ;; Or
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| 30 | (identical-points B C) ; B = C
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[99] | 31 | )
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| 32 | )
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