This question is similar to Chapter 1 Class 10 Real Numbers - Ex 1.1

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https://www.teachoo.com/1678/519/Ex-6.3--16---If-AD-and-PM-are-medians-of-triangles-ABC--PQR/category/Ex-6.3/

 

Question 37 (iii) (B) - If AM and DN are medians of triangles ABC, DEF - CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard

part 2 - Question 37 (iii) (B) - CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10
part 3 - Question 37 (iii) (B) - CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10 part 4 - Question 37 (iii) (B) - CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards - Class 10

 

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Question 37 (iii) (B) If AM and DN are medians of triangles ABC and DEF respectively then prove that ā–³ ABM ∼ ā–³ DEN.Given AM & DN are medians Since AM is median, M is mid-point of BC ∓ BM = CM = šŸ/šŸ BC Also, DN is median, N is mid-point of EF ∓ EN = FN = šŸ/šŸ EF We are assuming ā–³ ABC ∼ ā–³ DEF Otherwise this question cannot be solved Since ā–³ ABC ∼ ā–³ DEF And, Sides of similar triangles are proportional So, š“šµ/š·šø=š“š¶/š·š¹=šµš¶/šøš¹ š“šµ/š·šø=šµš¶/šøš¹ Putting BC = 2BM, and EF = 2EN š“šµ/š·šø=2šµš‘€/2šøš‘ š‘Øš‘©/š‘«š‘¬=š‘©š‘“/š‘¬š‘µ Also, since ā–³ ABC ∼ ā–³ DEF And, corresponding angles of similar triangles are equal ∓ ∠ B = ∠ E Now, In Ī” ABM & Ī”DEN āˆ šµ=āˆ šø š“šµ/š·šø=šµš‘€/šøš‘ Hence by SAS similarly Ī”ABM ∼ Ī”DEN Hence proved

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