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  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Ex 6.6, 2 In Fig. 6.57, D is a point on hypotenuse AC of Δ 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, Δ BMD ∼ Δ DMC In ∆ BDA Right Angled at D, Given DN ⊥ AB Applying theorem 6.7, Δ AND ∼ Δ DNB Using equation (1) ∆ BMD ∼ ∆DMC BM/𝐷𝑀 = MD/𝑀𝐶 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= 𝐷𝑁/𝐵𝑁 AN/DN= 𝐷𝑁/𝐵𝑁 AN . BN = (DN)2 (DN)2 = AN . BN Putting BN = DM from (4) (DN)2 = AN . DM Hence proved

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