Last updated at Dec. 8, 2016 by Teachoo

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Ex 6.3, 10 CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ∼ ΔFEG, show that: 𝐶𝐷/𝐺𝐻 = 𝐴𝐶/𝐹𝐺 Given: Δ ACB and Δ EGF respectively And , CD is the bisectors of ∠ ACB ∠ ACD = ∠ BCD = 1/2 ∠ ACB & GH is the bisectors of ∠ EGF ∠ FGH = ∠ EGH = 1/2 ∠ EGF & ΔABC ∼ ΔFEG To Prove: 𝐶𝐷/𝐺𝐻 = 𝐴𝐶/𝐹𝐺 Proof: Given ΔABC ∼ ΔFEG We know that corresponding angles of similar triangles are equal So, ∠A = ∠F And , ∠C = ∠G From (2) ∠C = ∠G 1/2 ∠C = 1/2∠G ∠ ACD = ∠ FGH In Δ ACD & Δ FGH ∠A = ∠F ∠ ACD = ∠ FGH Hence by AA similarity Δ ACD ∼ Δ FGH We know that corresponding sides of similar triangles are in proportion 𝐶𝐷/𝐺𝐻 = 𝐴𝐶/𝐹𝐺 Hence proved Ex 6.3, 10 CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ∼ ΔFEG, show that: (ii) ΔDCB ∼ ΔHGE Given ΔABC ∼ ΔFEG We know that corresponding angles of similar triangles are equal So, ∠B = ∠E And , ∠C = ∠G From (2) ∠C = ∠G 1/2 ∠C = 1/2∠G ∠ BCD = ∠ EGH In Δ CDB & Δ HGE ∠B = ∠E ∠BCD = ∠ EGH Hence by AA similarity Δ CDB ∼ Δ HGE Hence proved Ex 6.3, 10 CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ∼ ΔFEG, show that: (iii) ΔDCA ∼ ΔHGF Given ΔABC ∼ ΔFEG We know that corresponding angles of similar triangles are equal So, ∠A = ∠F And , ∠C = ∠G From (2) ∠C = ∠G 1/2 ∠C = 1/2∠G ∠ ACD = ∠ FGH In Δ ACD & Δ FGH ∠A = ∠F ∠ ACD = ∠ FGH Hence by AA similarity Δ ACD ∼ Δ FGH Hence proved

Chapter 6 Class 10 Triangles

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .