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Ex 6.4, 3 - ABC and DBC are two triangles on same base BC - Area of similar triangles

  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Ex 6.4, 3 In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that (π‘Žπ‘Ÿ(𝐴𝐡𝐢))/(π‘Žπ‘Ÿ(𝐷𝐡𝐢)) = 𝐴𝑂/𝐷𝑂 Given: βˆ† ABC and βˆ† DBC Having common base BC To prove: (π‘Žπ‘Ÿ βˆ† 𝐴𝐡𝐢)/(π‘Žπ‘Ÿ βˆ† 𝐷𝐡𝐢)=𝐴𝑂/𝐷𝑂 Proof: Since AO and OD are not part of βˆ† 𝐴𝐡𝐢 and βˆ† 𝐷𝐡𝐢 , we cannot directly use the theorem We know that Area of triangle = 1/2Γ—π΅π‘Žπ‘ π‘’Γ—π΄π‘™π‘‘π‘–π‘‘π‘’π‘‘π‘’ Lets draw altitude AE [ AE βŠ₯𝐡𝐢 ] Hence, ar βˆ† ABC=1/2×𝐡𝐢×𝐴𝐸 Similarly, for Ξ” DBC Let us draw altitude DF (DF βŠ₯ BC) Hence, ar βˆ† DBC=1/2×𝐡𝐢×𝐷𝐹 Now, taking ratio (π‘Žπ‘Ÿ βˆ†π΄π΅πΆ)/(π‘Žπ‘Ÿ βˆ†π·π΅πΆ) = (1/2 Γ— 𝐡𝐢 Γ— 𝐴𝐸)/(1/2 Γ— 𝐡𝐢 Γ— 𝐷𝐹) (π‘Žπ‘Ÿ βˆ†π΄π΅πΆ)/(π‘Žπ‘Ÿ βˆ†π·π΅πΆ) = 𝐴𝐸/𝐷𝐹 Now in βˆ† AOE and Ξ” DOF ∠ AEO = ∠ DFO ∠ AOE = ∠ DOF So, by using AA similarity criterion βˆ† AOE ~ βˆ† DOF If two triangle are similar their , corresponding sides are in the same ratio So, 𝐴𝐸/𝐷𝐹=𝐴𝑂/𝐷𝑂 Putting (2) in (1) (π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ βˆ† 𝐴𝐡𝐢)/(π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ βˆ† 𝐷𝐡𝐢)=𝐴𝐸/𝐷𝐹 (π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ βˆ† 𝐴𝐡𝐢)/(π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ βˆ† 𝐷𝐡𝐢)=𝐴𝑂/𝐷𝑂 Hence proved

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