Question

In the given figure, O is the centre of the circle and L and M are the mid points of AB and C B respectively .If angle OAB = angle OCB prove that BL = BM.

Answer 1

Assuming that AB and BC are chords in the circle(See the attached figure)

Angle OAB = angle OCB     --- given
Angle BLO = angle BMO= 90 --- Property of a chord which states that the line joining the center of the circle to the center of the chord is perpendicular to the chord ( in other words perpendicular bisectors of 2 chords of a circle, intersect at its center)
Side OB = Side OB            ---- common side and radius of the circle

Hence Δ BLO ≡ Δ BMO      - by AAS (angle angle side)

Hence Side BL = BM          - Corresponding parts of congruent triangles are congruent

Answer 2

GIVEN: ANGLE OAB= ANGLE OCB
​              L AND M ARE MID POINTS OF AB AND BC
CONSTRUCTION: JOIN A TO C
                                  JOIN  O TO L AND O TO M
TO PROVE: BL=BM
 
PROOF:
                OL IS PERP. BISECTOR OF AB AND OM IS PERP. BISECTOR OF BC
               SO, ANGLE OLA = ANGLE OMC
              
IN TRIANGLE. ALO AND CMO
AO=CO (RADII OF CIRCLE)
ALO= CMO ( 90 EACH)
LAO=MCO  (GIVEN)
BY AAS 
       TRI. ALO = TRI. CMO
      BY CPCT 
                      OL=OM
SINCE OL =OM,
                     AB =BC (EQIDISTANT CHORDS ARE EQUAL)
    1/2AB=1/2BC
​i.e. BL=CM
       HENCE PROVED