Basic Proportionality Theorem or Thales Theorem

Basic Proportionality Theorem or Thales Theorem

Statement: 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.

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Converse of Basic Proportionality Theorem

Statement: 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 DABC and a line l intersecting AB in D and AC in E,

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Basic Proportionality Theorem Example Problems With Solutions

Example 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

(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.

Solution:

Basic-Proportionality-Theorem-or-Thales-Theorem-Example-1

Example 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.

Solution:    In ΔTXM, we have

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Example 3:    In fig., EF || AB || DC. Prove that (frac{AE}{ED}=frac{BF}{FC}).

Solution:    We have, EF || AB || DC

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Example 4:    In figure, ∠A = ∠B and DE || BC. Prove that AD = BE

Solution:

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Example 5:    In fig., DE || BC. If AD = 4x – 3, DB = 3x – 1, AE = 8x – 7 and EC = 5x – 3, find the value of x.

Solution:

Basic-Proportionality-Theorem-or-Thales-Theorem-Example-5

Example 6:    Prove that the line segment joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.

Solution:    

Given : A quadrilateral ABCD in which P, Q, R, S are the midpoints of AB, BC, CD and DA respectively.

To prove: PQRS is a parallelogram.

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Example 7:    In fig. DE || BC and CD || EF. Prove that AD2 = AB × AF.

Solution:

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Example 8:    Ex.8 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 (frac{1}{x}+frac{1}{y}=frac{1}{z}).

Solution:

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Example 9:    In fig., LM || AB. If AL = x – 3, AC = 2x, BM = x – 2 and BC = 2x + 3, find the value of x.

Solution:

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Example 10:    In a given ∆ABC, DE || BC and (frac{AD}{DB}=frac{3}{4}). If AC = 14 cm, find AE.

Solution:

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Example 11:    In figure, DE || BC. Find AE.

Solution:

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Example 12:    In figure, ABC is a triangle in which AB = AC. Points D and E are points on the sides AB and AC respectively such that AD = AE. Show that the points B, C, E and D are concyclic.

Solution:    In order to prove that the points B, C, E and D are concyclic, it is sufficient to show that

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Example 13:    In fig., (frac{AD}{DB}=frac{1}{3}text{   and   }frac{AE}{AC}=frac{1}{4}). Using converse of basic proportionality theorem, prove that DE || BC.

Solution:

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Example 14:    Using basic proportionality theorem, prove that the lines drawn through the points of trisection of one side of a triangle parallel to another side trisect the third side.

Solution:

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Example 15:    In the given figure, (frac{AD}{DB}=frac{AE}{EC}) and ∠ADE = ∠ACB. Prove that ∆ABC is an isosceles triangle.

Solution:

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Example 16:    In fig., if DE || AQ and DF || AR. Prove that EF || QR.

Solution:

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Example 17:    Two triangles ABC and DBC lie on the same side of the base BC. From a point P on BC, PQ || AB and PR || BD are drawn. They meet AC in Q and DC in R respectively. Prove that QR || AD.

Solution:    Given: Two triangles ABC and DBC lie on the same side of the base BC. Points P, Q and R are points on BC, AC and CD respectively such that PR || BD and PQ || AB.

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Example 18:    ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF || AB. Show that (frac{AE}{ED}=frac{BF}{FC})

Solution:     Given: A trap ABCD in which AB || DC.

E and F are points on AD and BC respectively such that EF || AB.

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Example 19:    In fig., A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:   

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Example 20:    Any point X inside ∆DEF is joined to its vertices. From a point P in DX, PQ is drawn parallel to DE meeting XE at Q and QR is drawn parallel to EF meeting XF in R. Prove that PR || DF.

Solution:    A ΔDEF and a point X inside it. Point X is joined to the vertices D, E and F. P is any point on DX. PQ || DE and QR || EF.

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Thus, in ΔXFD, points R and P are dividing sides XF and XD in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have, PR || DF

Example 21:    Prove that any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.

Solution:    Given: A trapezium ABCD in which DC || AB and EF is a line parallel to DC and AB.

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Example 22:    Prove that the line drawn from the mid-point of one side of a triangle parallel of another side bisects the third side.

Solution:    Given: A DABC, in which D is the mid-point of side AB and the line DE is drawn parallel to BC, meeting AC in E.

To Prove : E is the mid-point of AC i.e.,

AE = EC.

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Hence, E bisects AC.

Example 23:    Prove that the line joining the mid-point of two sides of a triangle is parallel to the third side.

Solution:    Given: A ΔABC in which D and E are mid-point of sides AB and AC respectively.

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Thus, the line DE divides the sides AB and AC of ΔABC in the same ratio. Therefore, by the converse of Basic Proportionality Theorem, we have

DE || BC

Example 24:    AD is a median of ∆ABC. The bisector of ∠ADB and ∠ADC meet AB and AC in E and F respectively. Prove that EF || BC.

Solution:    Given: In ∆ABC, AD is the median and DE and DF are the bisectors of ∠ADB and ∠ADC respectively, meeting AB and AC in E and F respectively.

To Prove: EF || BC

Proof: In ∆ADB, DE is the bisector of ∠ADB.

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Thus, in ∆ABC, line segment EF divides the sides AB and AC in the same ratio.

Hence, EF is parallel to BC.

Example 25:    O is any point inside a triangle ABC. The bisector of ∠AOB, ∠BOC and ∠COA meet the sides AB, BC and CA in point D, E and F respectively. Show that AD × BE × CF = DB × EC × FA.

Solution:    

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