Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
9.9k views
in Circles by (44.3k points)

In the given figure, ABCD is a quadrilateral in which AD = BC and ∠ADC = ∠BCD. Show that the points A, B, C, D lie on a circle.

1 Answer

0 votes
by (58.8k points)
selected by
 
Best answer

It is given that ABCD is a quadrilateral in which AD = BC and ∠ADC = ∠BCD

Construct DE ⊥ AB and CF ⊥ AB

Consider △ ADE and △ BCF

We know that

∠AED + ∠BFC = 90o

From the figure it can be written as

∠ADE = ∠ADC – 90o = ∠BCD – 90o = ∠BCF

It is given that

AD = BC

By AAS congruence criterion

△ ADE ≅ △ BCF

∠A = ∠B (c. p. c. t)

We know that the sum of all the angles of a quadrilateral is 360o

∠A + ∠B + ∠C + ∠D = 360o

By substituting the values

2 ∠B + 2 ∠D = 360o

By taking 2 as common

2 (∠B + ∠D) = 360o

By division

∠B + ∠D = 360/2

So we get

∠B + ∠D = 180o

So, ABCD is a cyclic quadrilateral.

Therefore, it is proved that the points A, B, C and D lie on a circle.

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

...