Tiêu đề 02-Similar Triangles (Part-1)(1).pdf ✅

SIMILAR TRIANGLES 2 CHAPTER CONTENTS • Concept of Similarity • Thales Theorem • Criteria for Similarity of Triangles • Area of Two Similar Triangles • Phythagoras Theorem • Some Important Theorems ➢ CONCEPT OF SIMILARITY Geometric figures having the same shape but different sizes are known as similar figures. Two congruent figures are always similar but similar figures need not be congruent. Illustration 1 : Any two line segments are always similar but they need not be congruent. They are congruent, if their lengths are equal. Illustration 2 : Any two circles are similar but not necessarily congruent. They are congruent if their are equal. Illustration 3 : (i) Any two square are similar (see fig. (i)) Fig.(i) Fig.(ii) (ii) Any two equilateral triangles are similar (see fig. (ii)) ➢ SIMILAR POLYGONS  Definition Two polygons are said to be similar to each other, if (i) their corresponding angles are equal, and (ii) the lengths of their corresponding sides are proportional. If two polygons ABCDE and PQRST are similar, then from the above definition it follows that : Angle at A = Angle at P, Angle at B = Angle at Q, Angle at C = Angle at R, Angle at D = Angle at S, Angle at E = Angle at T and, PQ AB = QR BC = RS CD = ST DE If two polygons ABCDE and PQRST, are similar, we write ABCED ~ PQRST. Here, the symbol ‘~’ stands for is similar to. D C A B E 4 cm 3 cm 3 cm 4 cm 3.2 cm 100° 90° 120° 130° S R P Q T 6 m 100° 90° 120° 130° 6 cm 4.5 cm 4.8 cm 4.5 cm
➢ SIMILAR TRIANGLE AND THEIR PROPERTIES  Definition Two triangles are said to be similar, if their (i) corresponding angles are equal and, (ii) corresponding sides are proportional. Two triangles ABC and DEF are similar, if (i) A = D, B = E, C = F and, (ii) DE AB = EF BC = DF AC B C A D E F ➢ SOME BASIC RESULTS ON PROPORTIONALITY  Basic Proportionality Theorem or Thales Theorem If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. Given : A triangle ABC in which DE || BC, and intersects AB in D and AC in E. To Prove : DB AD = EC AE Construction : Join BE, CD and draw EF ⊥ BA and DG ⊥ CA. B C E G A F D Proof : Since EF is perpendicular to AB. Therefore, EF is the height of triangles ADE and DBE. Now, Area(ADE) = 2 1 (base × height) = 2 1 (AD.EF) and, Area(DBE) = 2 1 (base × height) = 2 1 (DB.EF)  Area( DBE) Area( ADE)   = (DB.BF) 2 1 (AD.EF) 2 1 = DB AD .... (i) Similarly, we have Area( DEC) Area( ADE)   = (EC.DG) 2 1 (AE.DG) 2 1 = EC AE .... (ii) But, DBE and DEC are on the same base DE and between the same parallels DE and BC.  Area (DBE) = Area (DEC)  Area( DBE) 1  = Area( DEC) 1  [Taking reciprocals of both sides]  Area ( DBE) Area ( ADE)   = Area ( DEC) Area ( ADE)   [Multiplying both sides by Area (ADE)]  DB AD = EC AE [Using (i) and (ii)] Corollary : If in a ABC, a line DE || BC, intersects AB in D and AC in E, then : (i) AD AB = AE AC (ii) DB AB = EC AC Proof : (i) From the basic proportionality theorem, we have DB AD = EC AE  AD DB = AE EC [Taking reciprocals of both sides]  1 + AD DB = 1 + AE EC [Adding 1 on both sides]  AD AD + DB = AE AE + EC  AD AB = AE AC (ii) From the basic proportionality theorem, we have DB AD = EC DE
 DB AD + 1 = EC AE + 1 [Adding 1 on both sides]  DB AD + DB = EC AE + EC  DB AB = EC AC So, if in a ABC, DE || BC, and intersect AB in D and AC in E, then we have (i) DB AD = EC AE (ii) AD DB = AE EC (iii) AD AB = AE AC (iv) AB AD = AC AE (v) DB AB = EC AC (vi) AB DB = AC EC  Converse of Basic Proportionality Theorem If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side. Given : A ABC and a line l intersecting AB in D and AC in E, such that DB AD = EC AE To prove :  || BC i.e. DE || BC B C D E F  A Proof : If possible, let DE be not parallel to BC. Then, there must be another line parallel to BC. Let DF || BC. Since DF || BC. Therefore from Basic Proportionality Theorem, we get DB AD = FC AF .... (i) But, DB AD = EC AE (Given) ....(ii) From (i) and (ii), we get FC AF = EC AE  FC AF + 1 = EC AE + 1 [Adding 1 on both sides]  FC AF + FC = EC AE + EC  FC AC = EC AC  FC = EC This is possible only when F and E coincide i.e. DF is the line l itself. But, DF || BC. Hence, l || BC. ❖ EXAMPLES ❖ Ex.1 D and E are points on the sides AB and AC respectively of a ABC such that DE || BC. Find the value of x, when D E B C A (i) AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm (ii) AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x cm. Sol.(i) In ABC, DE || BC  DB AD = EC AE (By thales theorem) x 4 4 − = 3x 19 8 − 4(3x – 19) = 8(x – 4) 12x – 76 = 8x – 32 4x = 44 x = 17 (ii) In ABC, DE || BC  DB AD = EC AE (By thales theorem) 3x 4 7x 4 + − = 3x 5x − 2 21x2 – 12x = 15x2 – 6x + 20x – 8 6x2 – 26x + 8 = 0 3x2 – 13x + 4 = 0 (x – 4) (3x – 1) = 0  x = 4, 1/3
Ex.2 Let X be any point on the side BC of a triangle ABC. If XM, XN are drawn parallel to BA and CA meeting CA, BA in M, N respectively; MN meets BC produced in T, prove that TX2 = TB × TC. Sol. In TXM, we have M X T N B C A XM || BN  TX TB = TN TM .... (i) In TMC, we have XN || CM  TC TX = TM TN .... (ii) From equations (i) and (ii), we get TX TB = TC TX  TX2 = TB × TC Ex.3 In fig., EF || AB || DC. Prove that ED AE = FC BF . D C P E F A B Sol. We have, EF || AB || DC  EP || DC Thus, in ADC, we have EP || DC Therefore, by basic proportionality theorem, we have ED AE = PC AP .... (i) Again, EF || AB || DC  FP || AB Thus, in CAB, we have FP || BA Therefore, by basic proportionality theorem, we have FC BF = PC AP ....(ii) From (i) and (ii), we have ED AE = FC BF Ex.4 In figure, A = B and DE || BC. Prove that AD = BE Sol. C D E A B A = B (given)  BC = AC ....(i) (Sides opposite to equal angles are equal) Now, DE || AB  DA CD = EB CE (By basic proportionality theorem)  DA CD + 1 = EB CE + 1 (Adding 1 on both sides)  DA CD + DA = EB CE + EB  DA CA = EB CE  AD AC = BE BC  AD AC = BE AC  AD 1 = BE 1  AD = BE