Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

 

Using the above theorem prove that a line through the point of intersection of the diagonals and parallel to the base of the trapezium divides the non parallel sides in the same ratio.

 

This question is Similar to Theorem 6.1 - Chapter 6 Class 10 - Triangles

This question is Similar to Ex 6.2, 9 - Chapter 6 Class 10 - Triangles

 

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Question 33 Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio. Using the above theorem prove that a line through the point of intersection of the diagonals and parallel to the base of the trapezium divides the non parallel sides in the same ratio.Let’s prove the theorem first Theorem If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ratio. Given: Δ ABC where DE ∥ BC To Prove: 𝐴𝐷/𝐷𝐵 = 𝐴𝐸/𝐸𝐶 Construction: Join BE and CD Draw DM ⊥ AC and EN ⊥ AB. Proof:ar (ADE) = 1/2 × Base × Height = 1/2 × AD × EN ar (BDE) = 1/2 × Base × Height = 1/2 × DB × EN Divide (1) and (2) "ar (ADE)" /"ar (BDE)" = (1/2 " × AD × EN" )/(1/2 " × DB × EN " ) "ar (ADE)" /"ar (BDE)" = "AD" /"DB" ar (ADE) = 1/2 × Base × Height = 1/2 × AE × DM ar (DEC) = 1/2 × Base × Height = 1/2 × EC × DM Divide (3) and (4) "ar (ADE)" /"ar (DEC)" = (1/2 " × AE × DM" )/(1/2 " × EC × DM " ) "ar (ADE)" /"ar (DEC)" = "AE" /"EC" Now, ∆BDE and ∆DEC are on the same base DE and between the same parallel lines BC and DE. ∴ ar (BDE) = ar (DEC) Hence, "ar (ADE)" /"ar (BDE)" = "ar (ADE)" /"ar (DEC)" "AD" /"DB" = "AE" /"EC" Hence Proved Now, let’s look at our questionUsing the above theorem prove that a line through the point of intersection of the diagonals and parallel to the base of the trapezium divides the non parallel sides in the same ratio. Given: ABCD is a trapezium where diagonals AC & BD intersect at O EF is a line passing through O, parallel to CD To prove: We need to prove 𝑨𝑬/𝑬𝑫=𝑩𝑭/𝑭𝑪 Proof: Using theorem If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ratio In Δ ADB Since EO ∥ CD Using Basic Proportionality theorem 𝑨𝑬/𝑫𝑬=𝑩𝑶/𝑫𝑶 In Δ BDC Since OF ∥ CD Using Basic Proportionality theorem 𝑩𝑶/𝑫𝑶=𝑩𝑭/𝑭𝑪 Comparing (1) and (2) 𝑨𝑬/𝑬𝑫=𝑩𝑭/𝑭𝑪 Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.