BPT or Thales Theorem

MATHS TRIANGLE BPT OR THALES THEOREM BPT OR THALES THEOREM Statement : If a line is drawn parallel to one side of a triangle to intersect the other sides in distinct points, then the other two sides are divided in the same ratio. Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively. To Prove : AD= AE DB EC Construction : Join BE and CD and draw DM ⊥ AC and EN ⊥ AB. Proof : 1 2 Area of ADE (= base × height) = 1 AD × EN. 2 Area of ADE is denoted as are (ADE) So, ar(ADE) = 1 DB × EN 2 And ar(BDE) = 1 DB × EN , 2 1 ar(ADE) 2 ADEN AD = = Therefore, ar(BDE) 1 DBEN DB 2 1 2 ….(i) 1 2 Similarly, ar(ADE = AE × DM and ar(DEC = EC × DM. 1 MATHS And 1 ar(ADE) 2 AEDM AE = = ar(DEC) 1 ECDM EC 2 ......(ii) Note that  BDE and  DEC are on the same base DE and between the two parallel lines BC and DE. So, ar(BDE) = ar(DEC) .....(iii) Therefore, from (i), (ii) and (iii), we have : AD= AE DB EC Hence Proved. COROLLARY If in a ABC, a line DE || BC, intersects AB in D and AC in E, then (i) DB= EC AD AE (ii) AB= AC AD AE (iii) AD= AE AB AC (iv) AB= AC DB EC (v) DB= EC AB AC Ex.1: In DABC, DE is parallel to BC and intersects AB and AC at D and E respectively, then (i) Sol. AB = AC DC EC (ii) AB = AC AD AE Proof : (i) By proportionality Theorem AD+1= AE +1 DB EC 2 AD = AE On adding 1 to both sides DB EC MATHS   (ii) AD = AE DB EC AD+DB = AE+EC DB EC AB = AC DB EC (By basic proportionality Theorem) Taking inverse and then adding 1 to both sides 1+ DB =1+ EC AD AE AD+DB = AE+EC AD AE AB = AC AD AE CONVERSE OF BASIC PROPORTIONALITY THEOREM : Statement : If a line divides any two sides of a triangle in the same ratio, the line must be parallel to the third line. Given : A A triangle ABC and line l intersecting AB at D and AC at E such that AD = AE . DB EC F D To Prove : DE || BC Proof : Let us suppose that DE is not parallel to BC. Then, B through D there must be some other line DF (let) parallel to BC. since DF || BC, by basic proportionality theorem, we get AD = AF DB FC AD = AE But DB EC From (1) and (2), .....(1) (given) .....(2) AF = AE On adding 1 to both sides FC EC AF +1= AE +1 FC EC 3 E l C MATHS AF +FC = AE+EC FC EC AC = AC FC EC Hence, FC = EC But this is impossible unless the points F and E coincide, i.e. DF and DE are coincident lines. Hence, DE || BC. Ex.2 LM || AB. If AL = x – 3, AC = 2x, BM = x – 2, BC = 2x + 3. find the value of x? Sol. In D ABC we have LM || AB AL = BM LC MC AL = BM AC− AL BC−BM x−3 = x−2 2x−(x−3) (2x+3) −(x−2) C L x−3 = x−2 x+3 x+5 A (x – 3)(x + 5) = (x – 2)(x + 3) x2 + 2x – 15 = x2 + x – 6 x=9 Ex.3 In the given figure PA, QB and RC each is perpendicular to AC such that PA = x, RC = y, QB = z, AB = a and BC = b. Prove that P R Q x y z A Sol. a B b C PA ⊥ AC and QB ⊥ AC  QB || PA 4 1+1 =1 x y z M B MATHS Thus in D PAC, QB || PA  QBC ~  PAC so QB = BC PA AC z= b x a+b  ..... (i) (by the property of similar triangle) In  RAC, QB || RC, so  QAB ~  RAC QB = AB RC AC z= a y a+b  ..... (ii) (by the property of similar triangle) Now from (i) and (ii) we get z + z = b + a  =1 x y  a+b a+b z z  + =1 x y  Ex.4 1+1 =1 x y z 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 A D E B C (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  AD= AE (By thales theorem) DB EC 5 MATHS 4 = 8 x −4 3x −19 4(3x – 19) = 8(x – 4) 12x – 76 = 8x – 32 4x = 44 x = 17 (ii) In ABC, DE || BC  AD= AE DB EC (By thales theorem) 7x −4 = 5x −2 3x +4 3x 21x2 – 12x = 15x2 – 6x + 20x – 8 6x2 – 26x + 8 = 0 3x2 – 13x + 4 = 0 (x – 4) (3x – 1) =   x = 4, 1/3 6