[Class 10] In the given figure ∠CEF = ∠CFE. F is the midpoint of DC.

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

Question 33 (b) - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

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

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