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

  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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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 ΔABC ∼ ΔPQR. Given: ΔABC and ΔPQR AD is the median of Δ ABC ,PM is the median of Δ PQR 𝐴𝐵/𝑃𝑄=𝐴𝐶/𝑃𝑅=𝐴𝐷/𝑃𝑀 To Prove:- ΔABC ∼ ΔPQR. 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 𝐴𝐵/𝑃𝑄=𝐴𝐶/𝑃𝑅=𝐴𝐷/𝑃𝑀 𝐴𝐵/𝑃𝑄=𝐵𝐸/𝑄𝐿=𝐴𝐷/𝑃𝑀 𝐴𝐵/𝑃𝑄=𝐵𝐸/𝑄𝐿=2𝐴𝐷/2𝑃𝑀 𝐴𝐵/𝑃𝑄=𝐵𝐸/𝑄𝐿=𝐴𝐸/𝑃𝐿 ∴ ΔABE ∼ ΔPQL ∴ ΔABE ∼ ΔPQL We know that corresponding angles of similar triangles are equal. ∴ ∠BAE = ∠QPL Similarly, we can prove that ΔAEC ∼ ΔPLR We know that corresponding angles of similar triangles are equal. ∠CAE = ∠RPL Adding (4) & (5), ∠BAE + ∠CAE = ∠QPL + ∠RPL ∠CAB = ∠RPQ In ΔABC and ΔPQR, 𝐴𝐵/𝑃𝑄=𝐴𝐶/𝑃𝑅 ∠CAB = ∠RPQ ∴ ΔABC ∼ ΔPQR Hence proved

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