Definition D1
Write [X] for the area of a figure X. Call one side of a triangle the base, and the distance from the opposite vertex to the line through the base the height on that base.
Proof.
(1)That distance is the length of the perpendicular dropped from the vertex, and the foot of the perpendicular exists uniquely.A8, A2
(2)The area of a triangle and the lengths of its sides are greater than 0.A4
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Lemma L1
If three points X, Y, Z lie on a line ℓ with Y between X and Z, and W is not on ℓ, then [WXY] / [WYZ] = XY / YZ
Proof.
(1)The foot of the perpendicular from W to ℓ exists uniquely. Let h be the length of that perpendicular.assumption, A8
(2)The bases XY and YZ both lie on ℓ, so the height of △WXY on base XY and the height of △WYZ on base YZ are both h.(1), assumption, D1
(3)[WXY] = ½ · XY · h, [WYZ] = ½ · YZ · h(2), A1
(4)YZ, h, and [WYZ] are all greater than 0, so [WXY] / [WYZ] = XY / YZ(3), A4, A5, A6
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Lemma L2
In △ABC, if D is between A and B, E is between A and C, and DE is parallel to BC, then [DBE] = [DEC]
Proof.
(1)D is not on line BC, so line DE and line BC are two distinct parallel lines.assumption
(2)B and C lie on line BC, so the distance from B to line DE and the distance from C to line DE are equal. Call that value h′.(1), A3
(3)Taking DE as base, the heights of △DBE and △DEC are those two distances, hence both h′.(2), D1
(4)[DBE] = ½ · DE · h′ = [DEC](3), A1, E
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Theorem T
In △ABC, if D is between A and B, E is between A and C, and DE is parallel to BC, then AD / DB = AE / EC
Proof.
(1)A, D, B lie on line AB with D between them, and E is not on line AB.assumption
(2)Applying L1 to △ADE and △DBE gives [ADE] / [DBE] = AD / DB(1), L1
(3)A, E, C lie on line AC with E between them, and D is not on line AC, so in the same way [ADE] / [DEC] = AE / ECassumption, L1
(4)[DBE] = [DEC]L2
(5)AD / DB = [ADE] / [DBE] = [ADE] / [DEC] = AE / EC(2), (3), (4), E
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Corollary C
Under the same assumptions, AD / AB = AE / AC
Proof.
(1)D is between A and B, and E is between A and C, so AD + DB = AB, AE + EC = ACassumption, A7
(2)AD·EC = AE·DBT, A4, A5
(3)Adding AD·AE to both sides gives AD(AE + EC) = AE(AD + DB), that is, AD · AC = AE · AB(1), (2), A6, E
(4)AB and AC are greater than 0, so AD / AB = AE / AC(3), A4, A5
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