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Maharashtra State BoardSSC (English Medium) 10th Standard

◻ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE. - Geometry Mathematics 2

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Question

◻ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE. 

Sum
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Solution

Given:

◻ABCD is a parallelogram.

E is a point on side BC.

Line DE intersects ray AB in point T.

To prove: DE × BE = CE × TE

Proof: In ∆BET and ∆CED

∠BET = ∠CED   ...(Vertically opposite angles)

∠BTE = ∠CDE   ...(Alternate interior angles, AB || CD and DT is a transversal line)

By AA test of similarity,

∆BET ∼ ∆CED 

∴ `("BE")/("CE") = ("ET")/("ED")`

∴ DE × BE = CE × TE

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Chapter 1: Similarity - Practice Set 1.3 [Page 22]

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Balbharati Mathematics 2 [English] Standard 10 Maharashtra State Board
Chapter 1 Similarity
Practice Set 1.3 | Q 7 | Page 22
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