In the given figure ∠CEF = ∠CFE. F is the midpoint of DC. Prove that  AB/BD = AE/FD

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We know that Sides opposite to equal angles is equal Since ∠CEF = ∠CFE ∴ CE = CF And since F is mid-point of DC CF = DF So, we can write CE = CF = DF In Δ ABE We have to use Basic Proportionality Theorem If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ratio. Let’s draw DG ∥ BE By BPT 𝑨𝑫/𝑫𝑩=𝑨𝑮/𝑮𝑬 Adding 1 both sides 𝐴𝐷/𝐷𝐵+1=𝐴𝐺/𝐺𝐸+1 (𝐴𝐷 + 𝐷𝐵)/𝐷𝐵=(𝐴𝐺 + 𝐺𝐸)/𝐺𝐸 𝑨𝑩/𝑫𝑩=𝑨𝑬/𝑮𝑬 In Δ CDG We have to use Basic Proportionality Theorem If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, the other two sides are divided in the same ratio. Since FE ∥ DG By BPT 𝑪𝑭/𝑭𝑫=𝑪𝑬/𝑮𝑬 From (1): CE = CF = DF Our equation becomes 𝑪𝑭/𝑪𝑭=𝑪𝑬/𝑮𝑬 1=𝐶𝐸/𝐺𝐸 GE = CE Since GE = CE & CE = CF = FD ∴ GE = FD From (2) 𝐴𝐵/𝐷𝐵=𝐴𝐸/𝐺𝐸 Putting GE = FD from (3) 𝑨𝑩/𝑫𝑩=𝑨𝑬/𝑭𝑫 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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.