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ABCD is a parallelogram. Point P divides AB in the ratio 2:3 and point Q divides DC in the ratio 4:1. Prove that OC is half of OA

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Transcript

Since opposite sides of parallelogram are equal AB = CD Let AB = CD = x Point P divides AB in ratio 2 : 3 Thus, 𝑨𝑷/𝑩𝑷=𝟐/πŸ‘ Adding 1 both sides 𝐴𝑃/𝐡𝑃+1=2/3+1 (𝐴𝑃 + 𝐡𝑃)/𝐡𝑃=(2 + 3)/3 𝐴𝐡/𝐡𝑃=5/3 Since AB = x π‘₯/𝐡𝑃=5/3 BP = πŸ‘/πŸ“ 𝒙 Also, AP = 𝟐/πŸ“ 𝒙 Point P divides AB in ratio 2 : 3 Thus, 𝑨𝑷/𝑩𝑷=𝟐/πŸ‘ Adding 1 both sides 𝐴𝑃/𝐡𝑃+1=2/3+1 (𝐴𝑃 + 𝐡𝑃)/𝐡𝑃=(2 + 3)/3 𝐴𝐡/𝐡𝑃=5/3Point Q divides DC in ratio 4 : 1 Thus, 𝑫𝑸/𝑸π‘ͺ=πŸ’/𝟏 Adding 1 both sides 𝐷𝑄/𝑄𝐢+1=4+1 (𝐷𝑄 + 𝑄𝐢)/𝑄𝐢=5 𝐷𝐢/𝑄𝐢=5 In Ξ”AOP & Ξ”COQ ∠ AOP = ∠ COQ ∠ APO = ∠ CQO ∴ Ξ”AOP ~ Ξ”COQ Since sides of similar triangle are proportional Thus, 𝑨𝑷/π‘ͺ𝑸=𝑢𝑨/𝑢π‘ͺ Putting AP = 𝟐/πŸ“ 𝒙 and QC = 𝟏/πŸ“ 𝒙 (𝟐/πŸ“ 𝒙)/(𝟏/πŸ“ 𝒙)=𝑢𝑨/𝑢π‘ͺ 2/1=𝑂𝐴/𝑂𝐢 πŽπ‚=𝟏/𝟐 πŽπ€ ∴ OC is half of OA 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.