Question

In a circle with centre a chord AQ congurrent chord RS
Name, the two congerrent minor are
with reason​

Answer 1

Answer:

hey mate here is your answer..

Given two chords ab and CD of a circle C (O, r) OE perpendicular AB and OF perpendicular CD such that

OE

To prove:AB>CD

Construction :join OA and OC

PROOF we know that the perpendicular from the centre of a circle to a chord bisects the chord

AE=1/2 AB

And, CF=1/2 CD

Also, OA=OC=r. [2 nd]

And, OEOF>OE

OF SQUARE >OE SQUARE[3 rd]

now from the right triangle OEAand OFC we have;

OA square =OE square +AE square and OC square =OF square +CF square

OE square +AE square =OF square +OF square [OA=OC=r]

OA square +AE square =OF SQUARE +CF SQUARE >OE SQUARE +CF SQUARE [USING 3]

AE SQUARE >CD SQUARE

AB>CD

Hence, AB>CD

hope it helps you dear..

Step-by-step explanation: