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Ex 6.6, 2 (Optional) - D is a point on hypotenuse AC of triangle ABC

Ex 6.6, 2 (Optional) - Chapter 6 Class 10 Triangles - Part 2
Ex 6.6, 2 (Optional) - Chapter 6 Class 10 Triangles - Part 3 Ex 6.6, 2 (Optional) - Chapter 6 Class 10 Triangles - Part 4 Ex 6.6, 2 (Optional) - Chapter 6 Class 10 Triangles - Part 5 Ex 6.6, 2 (Optional) - Chapter 6 Class 10 Triangles - Part 6

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Ex 6.6, 2 In Fig. 6.57, D is a point on hypotenuse AC of ABC, DM BC and DN AB. Prove that : DM2 = DN . MC DN2 = DM . NA Given: ABC is a triangle and D is a point on hypotenuse AC such that BD AC DM BC And DN AB To Prove: (i) DM2 = DN . MC (ii) DN2 = DM . NA Proof: From Theorem 6.7 If a Perpendicular is drawn the vertex of the right angle of a right triangle to the hypotenuse, the triangle on both sides of the Perpendicular are similar to the whole triangle and to each other. In BDC Right angled at D Given DM BC Applying theorem 6.7, BMD DMC In BDA Right Angled at D, Given DN AB Applying theorem 6.7, AND DNB Using equation (1) BMD DMC BM/ = MD/ BM . MC = (DM)2 (DM)2 = BM . MC Now, we need to Prove BM = DN In ABC AB BC and DM BC AB DM NB DM Also, CB AB and DN AB CB DN MB DN Now, in quadrilateral DNBM ND DM & MB DN Since both pairs of opposite sides are parallel DNMB is a parallelogram Since opposite sides of parallelogram are equal DN = MB & DM = NB Putting BM = DN in equation (3) BM . MC = (DM)2 DN. MC = (DM)2 Putting BM = DN in equation (3) (DM)2 = BM . MC (DM)2 = DN . MC Hence proved. Now, we prove DN2 = DM . AN From (2) AND DNB AN/DN= / AN/DN= / AN . BN = (DN)2 (DN)2 = AN . BN Putting BN = DM from (4) (DN)2 = AN . DM 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.