Fewpal
0 votes
238 views
in Mathematics by (29.7k points)

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

1 Answer

0 votes
by (127k points)
selected by
 
Best answer

Given : A trapezium ABCD in which AB || CD and AD = BC. 

To Prove : ABCD is a cyclic trapezium. 

Construction : Draw DE ⊥ AB and CF ⊥ AB. 

Proof : In ∆DEA and ∆CFB, we have 

AD = BC [Given] 

∠DEA = ∠CFB = 90° [DE ⊥ AB and CF ⊥ AB]

DE = CF 

[Distance between parallel lines remains constant] 

∴ ∆DEA ≅ ∆CFB [RHS axiom] 

⇒ ∠A = ∠B ...(i) [CPCT] 

and, ∠ADE = ∠BCF ..(ii) [CPCT] 

Since, ∠ADE = ∠BCF [From (ii)] 

⇒ ∠ADE + 90° = ∠BCF + 90° 

⇒ ∠ADE + ∠CDE = ∠BCF + ∠DCF 

⇒ ∠D = ∠C ..(iii) 

[∠ADE + ∠CDE = ∠D, ∠BCF + ∠DCF = ∠C] 

∴ ∠A = ∠B and ∠C = ∠D [From (i) and (iii)] (iv) 

∠A + ∠B + ∠C + ∠D = 360° [Sum of the angles of a quadrilateral is 360°] 

⇒ 2(∠B + ∠D) = 360° [Using (iv)] 

⇒ ∠B + ∠D = 180° 

⇒ Sum of a pair of opposite angles of quadrilateral ABCD is 180°. 

⇒ ABCD is a cyclic trapezium Proved

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...