**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.

**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,

**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:**

**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 TX^{2} = TB × TC.

**Solution: **In ΔTXM, we have

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

**Solution: **We have, EF || AB || DC

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

**Solution:**

**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:**

**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.

**Example 7: **In fig. DE || BC and CD || EF. Prove that AD^{2} = AB × AF.

**Solution:**

**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:**

**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:**

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

**Solution:**

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

**Solution:**

**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

**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:**

**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:**

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

**Solution:**

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

**Solution:**

**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.

**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.

**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: **

**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.

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.

**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.

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.

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.

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: **