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The Area Concept , Similarity in Triangles, and Applications of the Basic Proportionality Theorem Section 4.3: The Area Concept (SMSG Postulates 17 – 20) The three properties which are necessary for all “area measure assignments”: 1) Every polygon is assigned a unique number called its “area”, 2) Congruent triangles (polygons) are assigned the same “area number”, 3) If a polygon P is the union of two regions R1 and R2 (which intersect only in a finite number of points and line segments), then Area of P = (Area of R1) + (Area of R2) Postulates 17, 18 and 19 ensure that the Euclidean Area Assignment meets requirements 1, 2 and 3, respectively. Postulate 20: The area of a rectangle is the product: Area = Base × Altitude. Theorem 4.3.1 (Proposition 35 in Book I of Euclid’s Elements): Two parallelograms that share a common base and whose sides opposite this base are contained in the same line are equal in area. Proof: Consider parallelograms ABCD and EBCF that share the common base and whose opposite sides are contained in the same line . A By the Converse of the Alternate Interior Angle Theorem, the angles indicated as congruent are congruent. D E F T2 G T1 B C Since the opposite sides of a parallelogram are congruent, the segments indicated as congruent are congruent. Now, AE = AD + DE = EF + DE = DE + EF = DF . Thus, and so, Δ ABE Δ DCF by ASA . Since two congruent figures have equal area measure, (Area of ΔABE) = (Area of ΔDCF) . Let Δ T1 = Δ BCG and Δ T2 = Δ DEG . Area of ABCD = (Area of ΔABE) + (Area of T1) – (Area of T2) and Area of EBCF = (Area of ΔDCF) + (Area of T1) – (Area of T2) . By substitution, Area of ABCD = (Area of ΔDCF) + (Area of T1) – (Area of T2) = Area of EBCF . QED 2 The Height of Triangles and Parallelograms (Relative to a chosen base) Definition: Given a triangle, choose one side to be called the base of the triangle. D C h h A B F1 C A F2 h B F3 The height h of the triangle (relative to the chosen base) is the length of the segment drawn perpendicular to the base from the vertex not on the base. (See below.) Given a parallelogram, choose one side to be called the base of the parallelogram. The height h of the parallelogram (relative to the chosen base) is the length of the segment drawn perpendicular to the base from any point on side opposite the base. (See above.) The Altitude of a Triangle (Relative to one of the sides) Definition: The term “altitude of a triangle” has 3 meanings depending on the context. Given Δ ABC, choose one side, say AB , with vertex C opposite the chosen side. 1. The line through C and perpendicular to the line containing AB is the altitude of the triangle drawn from C to the opposite side AB . 2. The line segment drawn from C to the foot of the perpendicular line, drawn from C to the line containing AB , is the altitude of the triangle drawn from C to the opposite side AB . 3. The length of the line segment drawn from C to the foot of the perpendicular line, drawn from C to the line containing AB , is the altitude of the triangle relative to side AB . In this sense, the altitude and the height of a triangle are the same. There are three altitudes of a triangle, each relative to one the three sides of the triangle. 3 Theorem 4.3.2, The "Parallelogram Area Formula" Theorem: The area of a parallelogram is the product of its height and the length of its base. Proof: Theorem is true for a rectangle (SMSG Postulate 20). D A C E B F These two parallelograms have the same height. They share the same base (Segment CD) and the sides opposite the base lie along the same line. By the previous theorem, they have the same area, which for the rectangle is: AREA = BASE × HEIGHT . Theorem 4.3.3: The area of a Right Triangle is: AREA = ½ (Base) (Height). C A' Δ ABC Δ A’CB by SSS Area(Δ ABC) = ½ Area(ٱABA’C) = ½ AB × AC A B = ½ (Base) (Height) 4 Theorem 4.3.4, The "Triangle Area Formula" Theorem: The area of a Triangle is: AREA = ½ (Base) (Height) . Proof: Every triangle has at least two acute angles. Assume that there are acute angles at A and B. Drop a perpendicular segment from C to D on side AB. C Area(Δ ABC) = Area(Δ ADC) + Area(Δ BDC) = ½ AD×CD + ½ DB×CD = ½ (AD + DB) × CD = A D B ½ (Base AB) (Height CD) Section 4.4: The Similarity Relation Definition: Two triangles Δ ABC and Δ A’B’C’ (or two n-gons …) are similar ( write Δ ABC ~ Δ A’B’C’ ) if there is a one-to-one correspondence between the vertices A ↔ A’ , B ↔ B’ , C ↔ C’ such that: 1) Corresponding angles are congruent, and 2) Any two pairs of corresponding sides are in the same proportion k, for some number k . Thus, k = ( A’B’ / AB ) = ( A’C’ / AC ) = ( B’C’ / BC ) . Theorem 4.4.1: Similarity of Polygons is an equivalence relation, i.e., it has the properties of Identity, Reflexivity and Transitivity. 5 Theorem 4.4.2 (The book's Basic Proportionality Theorem): If a line parallel to one side of a triangle intersects the other two sides in two different points, then it divides these sides into segments that are proportional. Given that B Line DE || Line AC, then, D E BD / BA = BE / BC and BD / DA = BE / EC. A C (Note: The first equality is proven in the book. The proof given in the textbook divides the area of the triangle into smaller triangular pieces in two different ways, and calculating the ratios of the areas these pieces in two ways results in the equality of the ratios of sides. The second equality follows from the first algebraically. It will also be the case that BD / BE = BA / BC , which also follows algebraically . ) Example: On the triangle to the right, it is given that BC = 15 and segment DE is parallel to segment AB. C 8 Determine x and y. Solution: AC = 11. By Theorem 4.4.2, D 8 / 11 = x / 15 x = (15)(8) / 11 = 10.91 3 A x / 8 = y / 3 = 10.91 / 8 x E y B y = (3)(10.91) / 8 = 4.09 Theorem 4.4.4 is the converse of this Theorem 4.4.2: If line DE cuts two sides of Δ ABC into proportional segments, then line DE is parallel to the third side of the triangle. 6 Similarity Conditions of Triangles and The Pythagorean Theorem Theorem 4.4.5 (The AAA Similarity Condition for Triangles): If, under a correspondence, the three interior angles of one triangle are congruent to the corresponding angles of a second triangle, then the triangles are similar, and so corresponding sides are proportional. Theorem 4.4.6 (SAS Similarity Condition for Triangles): If, under a correspondence, an angle of one triangle is congruent to the corresponding angle of another triangle, and if the corresponding sides that surround these angles are proportional, then the triangles are similar. Theorem 4.4.7 (SSS Similarity Condition for Triangles): If, under a correspondence, the lengths of the three sides of one triangle are proportional to the lengths of the corresponding sides of another triangle, then the triangles are similar. Theorem 4.4.8 (The Pythagorean Theorem): If a and b are the lengths of the legs of a right triangle and if c is the length of the hypotenuse, then a2 + b2 = c2 . 7 Applications of the Basic Proportionality Theorem: Angle Bisectors and Ratios of Segment Lengths Theorem (NIB) 4.1, The "Basic Proportionality Converse" Theorem, Theorem 4.4.2: (This is actually Theorem 4.4.4) Given ABC , let D be on AB , and E on AC such that AD AE DB EC or AD AE AB AC then DE is parallel to BC . Proof. Let F be the point where the line through D parallel to BC intersects AC . First, assume AD AE . AB AC A Thus, DF is parallel to BC . Thus, AE ( AC ) AD . AB D E F By the "Basic Proportionality" Theorem, AD AF . AB AC AD Thus, AF ( AC ) , and so, AE = AF by AB C B substitution. E = F by SMSG Postulates 3 and 4. Now, assume Thus DE = DF , which is parallel to BC . AD AE . As before, by the "Basic Proportionality" Theorem, DB EC AD AF AE AF . DB FC EC FC Since AC = AE + EC and AC = AF + FC , EC = AC - AE and FC = AC - AF . Thus, by substitution, AE AF , and so, (AE) (AC – AF) = (AF) (AC – AE) . AC AE AC AF (AE) (AC) – (AE) (AF) = (AF) (AC) – (AF) (AE) . Subtracting (AF) (AE) from both sides, we conclude that (AE) (AC) = (AF) (AC) . Dividing by (AC), AE = AF, and as before, E = F. Thus DE = DF , DE is parallel to BC . QED 8 Theorem (NIB) 4.2, The "Angle Bisector Third Side Cut" Theorem. The bisector of the internal angle of a triangle divides the opposite side in the ratio of the adjacent sides. That is, for a triangle ABC , if D is the point on AC such that BD is the bisector of ABC in ABC , then Proof. Suppose BD bisects ABC in ABC . = . At C, construct a line parallel to BD , intersecting AB at E. Thus, E . || 4 Let ∠ 1 , ∠ 2 , ∠ 3 , and ∠ 4 be the angles as indicated in the figure. Since BD is an angle bisector, ∠ 1 ∠ 2 . Since ∠ 1 and ∠ 4 are corresponding angles formed when intercepts parallel lines and , ∠1 ∠4. By transitivity, ∠ 2 ∠ 4 . B 1 2 Since ∠ 2 and ∠ 3 are alternate interior angles formed when intercepts parallel lines and , ∠ 2 ∠ 3 (AIA Converse). By transitivity, ∠ 3 ∠ 4 . By the "Congruent Base Angles" Theorem, BE = BC . By the "Basic Proportionality Theorem" applied to AEC , But, BE = BC, so = . 3 A D = C . QED Theorem (NIB) 4.3, The "Angle Bisector Third Side Cut Converse" Theorem: The converse of the "Angle Bisector Third Side Cut Converse" Theorem is true; that is, if D is the point on AC in ABC such that AB AD , BC DC then BD is the angle bisector of ABC . Proof: The proof is left as an exercise. Example: Given that AC = 25 and that BD is the bisector of ABC in the ABC shown in the figure, determine the values of x and y. B Solution: By the "Angle Bisector Third Side Cut" Theorem, Theorem (NIB) 4.2, x 14 2 y 21 3 Since x y 25 , and so, x 2 y y 3 2 y . 3 5 y 25 , 3 and so y = 15 and x = 10. 21 14 A x D y C