Definition D1
Call one side BC of △ABC the base and write b for its length. Let H be the foot of the perpendicular from A to line BC, call the length of AH the height, and write h for it.
Proof.
(1)A is not on line BC, so the foot of the perpendicular H exists uniquely.A10
(2)Write [X] for the area of a figure X. Polygons have areas.A4, convention
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Lemma L1
In △XYZ with a right angle at Y, if YZ = p and YX = q, then [XYZ] = ½pq
Proof.
(1)∠Y is a right angle and the two sides enclosing it are YZ = p, YX = q.assumption
(2)There exists a rectangle XYZW having those two sides as adjacent sides.(1), A5
(3)Opposite sides are equal, so XY = ZW, YZ = WX(2), A6
(4)The diagonal XZ is common to the two triangles, so their three sides are respectively equal. △XYZ ≅ △ZWX(3), A7
(5)The triangles △XYZ and △ZWX made by the diagonal XZ do not overlap and fill the rectangle XYZW.(2), A8
(6)The two triangles have areas and are congruent, so [XYZ] = [ZWX](4), A4, A2
(7)[XYZW] = [XYZ] + [ZWX] = 2[XYZ](5), (6), A3, A14
(8)The adjacent sides of the rectangle are p and q, so [XYZW] = pq(2), (1), A1
(9)2[XYZ] = pq, hence [XYZ] = ½pq(7), (8), A14, E
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Theorem T
For a triangle ABC with base b and corresponding height h, [ABC] = ½bh
Proof.
(1)H is on line BC, so if H ≠ B then ∠AHB is a right angle, and if H ≠ C then ∠AHC is a right angle.D1, A11
(2)Either H = B, or H = C, or H, B, C are three distinct points on line BC and exactly one lies between the other two. These five cases exhaust everything.D1, A13
(3)Case 1. If H = B, then △ABC has a right angle at B with enclosing sides BC = b, BA = AH = h, so [ABC] = ½bh(1), (2), L1
(4)Case 2. The case H = C is the same as (3) with the names B and C exchanged.(3)
(5)Case 3. If H lies between B and C, then BH + HC = b(2), A12
(6)The segment AH divides △ABC into △ABH and △AHC, and the two triangles do not overlap and fill △ABC.(5), A9
(7)The two triangles have right angles at H with enclosing sides BH and h, and HC and h, respectively, so [ABH] = ½·BH·h, [AHC] = ½·HC·h(1), (5), L1
(8)[ABC] = [ABH] + [AHC] = ½(BH + HC)h = ½bh(5), (6), (7), A3, A14
(9)Case 4. If B lies between H and C, then HB + b = HC(2), A12
(10)The segment AB divides △AHC into △AHB and △ABC, and the two triangles do not overlap and fill △AHC.(9), A9
(11)△AHB and △AHC have right angles at H, so [AHB] = ½·HB·h, [AHC] = ½·HC·h(1), (9), L1
(12)[AHC] = [AHB] + [ABC], so [ABC] = ½·HC·h − ½·HB·h = ½(HC − HB)h = ½bh(9), (10), (11), A3, A14, E
(13)Case 5. The case where C lies between H and B is the same as (9)–(12) with the names B and C exchanged.(12)
(14)In all five cases, [ABC] = ½bh(3), (4), (8), (12), (13)
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