In the given figure, DEFG is a square and ∠BAC = 90°. Show that FG 2 = BG × FC

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Question 23 (OR 1st question) In the given figure, DEFG is a square and ∠BAC = 900 . Show that FG2 = BG × FC Now, DE ∥ GF, So, we can say DE ∥ BC For parallel lines DE and BC, transversal AB ∠ ADE = ∠ GBD (Corresponding angles) For parallel lines DE and BC, transversal AC ∠ AED = ∠ FCE (Corresponding angles) Now, In Δ ADE and Δ GBD ∠ ADE = ∠ GBD ∠ DAE = ∠ DGB Δ ADE ∼ Δ GBD (From (1)) (Both 90° ) (AA Similarity) In Δ ADE and Δ FEC ∠ AED = ∠ FCE ∠ DAE = ∠ EFC Δ ADE ∼ Δ FEC (From (2)) (Both 90° ) (AA Similarity) So, we can write Δ GBD ∼ Δ FEC (Triangles similar to same triangles are similar) Hence, sides of similar triangle are in proportion 𝐺𝐷/𝐹𝐶=𝐺𝐵/𝐹𝐸 GD × FE = FC × GB Since DEFG is a square GD = FE = FG So, our equation becomes FG × FG = FC × GB FG2 = FC × GB Hence proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.