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Ex 6.3, 14 - Sides AB, AC and median AD of a triangle ABC - SAS Similarity

Ex 6.3, 14 - Chapter 6 Class 10 Triangles - Part 2
Ex 6.3, 14 - Chapter 6 Class 10 Triangles - Part 3 Ex 6.3, 14 - Chapter 6 Class 10 Triangles - Part 4 Ex 6.3, 14 - Chapter 6 Class 10 Triangles - Part 5

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Ex 6.3, 14 Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ABC PQR. Given: ABC and PQR AD is the median of ABC ,PM is the median of PQR / = / = / To Prove:- ABC PQR. Proof: Let us extend AD to point D such that AD = DE and PM up to point L such that PM = ML Join B to E, C to E, & Q to L, and R to L We know that medians is the bisector of opposite side Hence, BD = DC Also, AD = DE Hence in quadrilateral ABEC, diagonals AE and BC bisect each other at point D. Therefore, quadrilateral ABEC is a parallelogram. AC = BE and AB = EC Similarly, we can prove that PQLR is a parallelogram PR = QL, PQ = LR Given that / = / = / / = / = / / = / =2 /2 / = / = / ABE PQL ABE PQL We know that corresponding angles of similar triangles are equal. BAE = QPL Similarly, we can prove that AEC PLR We know that corresponding angles of similar triangles are equal. CAE = RPL Adding (4) & (5), BAE + CAE = QPL + RPL CAB = RPQ In ABC and PQR, / = / CAB = RPQ ABC PQR Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.