Loading....
Coupon Accepted Successfully!

 

Converse of Theorem 2

Converse : The line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord.
Given : A chord PQ of a circle C(O, r) and L is the mid-point of PQ.
To Prove : OL
PQ
Construction : Join OP and OQ

 

Proof :
In ΔOLP and ΔOLQ
OP = OQ = r (radii of the same circle)
PL = QL (given)
OL = OL (common)
...  
ΔOLP ≅ ΔOLQ (by CPCT)
OLP = OLQ
But
OLP + OLQ = 180° (Linear Pair)
..OLP = OLQ = 90°
Hence, OL
PQ.
 

Example :

L and M are the mid-points of two equal chords AB and CD of a circle with centre O (see Fig.) Prove that ALM =CML.

Solution :

Since L is the mid-point of AB and M is the mid-point of CD, therefore OL AB and OM CD. ALO = CMO = 90°
... OL = OM (equal chords of a circle are equidistant)
...
OLM = OML                     (i)
Now,
ALM = OLA - OLM = 90° - OLM  
..OLM = 90° - ALM          (ii) 
and
CML = OMC - OML
              = 90° -
OML
...OML = 90° - CML           (iii)
From Eqs (i), (ii) and (iii) we get,
ALM = CML.





Test Your Skills Now!
Take a Quiz now
Reviewer Name