Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent

A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent

Answers (1)

Given: A, B, C are 3 points on a circle

Construction \Rightarrow Draw a perpendicular dissector of AB and AC and they meet at a point O. Then join OA, OB and OC.

To prove: perpendicular bisector of AB, BC and CA are concurrent.

Proof: In \triangleOEA and \triangleOEB

AE = BE [OL is ^ar bisector of AB]

OE = OE [Common side]

\angleAEO = \angleBEO [each 90°]

\therefore \triangleOEA \cong \triangleDEB [By SAS congruence Rule]

OA = OC

Similarly, \triangleOFA \cong \triangleOFC [by SAS congruence]

\therefore OB = OC = OA = t [Say]

So, we draw a perpendicular from O to the BC on the join

In \triangleOMB and \triangleOMC

OB = OC [Proved Above]

OM = OM [Common side]

\angleOMB = \angleOMC [90° each]

\therefore \triangleOMB \cong \angleOMC [by RHS congruence rule]

\Rightarrow BN = MC [by corresponding parts of the congruent triangle]

Hence, OM is the perpendicular bisector of side BC.

Hence OL, ON, and OM are concurrent.

Hence proved. 

Posted by

infoexpert24

View full answer

Similar Questions

Latest Question