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Chapter-11 MID POINT THEOREM MATH CLASS-9

Module Objectives State the mid-point theorem. Prove mid-point theorem logically. State the converse of mid-point theorem. Solve problems and riders based on the mid-point theorem.

Analysis of Mid-Point Theorem Construct the triangles as given below. D and E are the mid-points of the sides of AB and AC of the ∆ABC. (i) (ii) (iii) Measure the lengths of DE and BC and the magnitudes of ABC and ADE in each case and complete the table given in the following slide.

Analysis of Mid-Point Theorem FIG NO. DE BC ADE ABC i) ii) iii) With the help of the above table answer the following questions: What is the relation between the lengths of DE and BC? What is the relation between ADE and ABC ? What type of angles are ADE and ABC ? Hence what type of lines are DE and BC ? From the measurements you will observe that, DE is half of BC. i.e. DE = 1/ 2 (BC).The corresponding angles ADE and ABC are equal. Hence , DE || BC.

Proof of Mid-Point Theorem Theorem : The line joining the mid-points of any two sides of a triangle is parallel to and half of the third side. Data : In ∆ABC , D is the midpoint of AB and E is the midpoint of AC To Prove: (i) DE ||BC (ii) DE = ½ BC Analysis: On what basis can we prove lines are parallel ? (1) Corresponding or alternate angles are equal. (2) The lines form the opposite sides of a parallelogram.

Proof of Mid-Point Theorem We not only have to prove DE || BC but also DE = ½ BC .How can we do this ? This is possible by constructing a parallelogram which has BC and DF as its sides. How can we show that DBCF is a parallelogram? This is possible by proving DB = CF and DB ||CF By constructing CF || DB. We cannot prove that DB = CF directly.But we know DB = DA. How can we prove that DB = CF ? So we will prove DA = CF Hence prove DB = CF by proving ∆ ADE ≡ ∆ EFC. How can we prove that ∆ ADE ≡ ∆ EFC ? By applying A.S.A postulate.

Proof of Mid-Point Theorem Hence ,in order to prove DE || BC and DE = ½ BC logically, we need to a) Construct CF || AB b) Prove ∆ ADE ≡ ∆ EFC. c) Prove DBCF is a parallelogram. Theorem : The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is equal to half of it. Data : In ∆ ABC , D is the midpoint of AB and E is the midpoint of AC.

Proof of Mid-Point Theorem To prove : DE || BC and DE = ½ BC. Construction : From C, draw a line parallel to BA. Produce DE to meet this line at F. STATEMENT REASON In ∆ADE and ∆ EFC, AE = EC given AED = CEF Vertically opposite angles. DAE = ECF Alternate angles. DA || CF Hence , ∆ ADE ≡ ∆ EFC. By A.S.A postulate Hence , AD = CF and DE = EF Corresponding parts of congruent ∆s. But AD = BD given., D is the midpoint of AB Therefore, CF = BD From the above two statements Thus, DBCF is a parallelogram. Opposite sides of CF and BD are equal and parallel. So, DE || BC Opposite sides of a parallelogram are parallel.

Proof of Mid-Point Theorem STATEMENT REASON Now DE = EF ∆ADE ≡ ∆ CEF DE = ½ BC In parallelogram DBCF, DF = BC Therefore , DE || BC and DE = ½ BC. Hence , it is proved that, the line segment joining the mid points of any two sides of a triangle is parallel to the third side and is equal to half of it. Note: Prove this theorem by changing the construction as : Produce DE to F such that EF = DE. Converse of the theorem : The straight line drawn through the midpoint of one side of a triangle and parallel to another , bisects the third side. Activity: a) Prove the above statement logically. b) Verify it by practical method.

Examples Example 1 : In the given figure , AB = 8.4 cm ,PR = 5.0 and PQ = 4.8cm .Find the length of BC,CA and QR. If APR =45⁰, find ABC. Given : AB = 8.4cm PR = 5.0 cm PQ = 4.8 cm Solution : BC = 2 * PR (Theorem 3) (2) CA = 2 * PQ (Theorem 3) BC = 2 * 5 CA = 2 * 4.8 BC = 10 cm CA = 9.6 cm QR = ½ AB (Theorem 3) (4) B = APR = 45⁰ QR = ½ * 8.4 QR = 4.2 cm Therefore, PR || BC , corresponding angles are equal.

Examples Example 2 : Prove that the figure obtained by joining the mid-points of the adjacent Sides of a quadrilateral ABCD. To Prove: PQRS is a parallelogram. Construction : Join BD. Proof: In ∆ABD, PS || BD and PS = ½ BD {Since, line joining the midpoints of two sides of a ∆ is parallel to and half the third side.} In ∆BCD, QR || BD and QR = ½ BD Therefore, PS || QR, PS = QR. Hence PQRS is a parallelogram.

Do it Yourself In ∆ ABC , D is mid-point of AB and E is mid point of BC. Calculate : i) DE , if AC = 6.4 cm. ii) DEB , if ACB = 63⁰ 2.In ∆ PQR , A,B,C are the midpoints of sides PQ,QR and RP. If the perimeter of ∆ PQR = 32 cm ,what is the perimeter of ∆ ABC ? 3) In the figure, X and Y are the midpoints of AB and AC respectively. Given that BC = 6cm.AB = 5.4 cm and AC = 5.0 cm. Calculate : i) The perimeter of ∆ AXY. ii) The perimeter of trapezium BCYX.

Do it Yourself 4) In ∆ ABC ,’Q’ and ‘M’ are the midpoints of AB and AC respectively. P and L are the midpoints of AQ and AM respectively .If BC = 12 cm , find the length of PL. 5) Construct ∆ ABC, given BC = 7cm, AB = 8 cm and AC = 6cm . Mark D and E the midpoints of AB and AC respectively. Construct parallelograms DBCF and ECBG. 6) Prove that the four triangles obtained by joining the midpoints of the sides of a triangle are congruent. 7) Prove that the figure obtained by the joining the midpoints of the adjacent sides of a rhombus is a rectangle. 8) In the given figure, PQRS is a parallelogram . A and B are the midpoints of sides SP and SR respectively .Prove that the area of ∆APR is equal to area of ∆BPR.

Do it Yourself 9) Use the following figure to prove that : ( AB || PR || DC ) i) R is the mid-point of BC. ii) PR = ½ ( AB + DC ) 10) D,E and F are the mid-points of the sides AB,BC and CA of an isosceles ∆ ABC in which AB = BC. Prove that ∆ DEF is also isosceles.

END OF CHAPTER