Question

●In the given figure, AB and CD are the chords intersecting at O. OM, ON are the bisectors of /_AOD & /_BOC.

Prove that AM/DM=CN/BN
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Answer 1

Answer:

In ∆ OBN and ∆ OCN,

OB = OC

angle BON = angle CON

ON = ON

Thus, by SAS criterion of congruency of triangles,

∆ OBN =~ ∆ OCN

Hence, by CPCT,

BN = CN

Therefore, CN = BN ---- ( 1 )

Now, in ∆ OAM and ∆ ODM,

AO = DO

angle AOM = angle DOM

OM = OM

Thus, by SAS criterion of congruency of triangles,

∆ OAM =~ ∆ ODM

Hence, by CPCT,

AM = DM --- ( 2 )

From ( 1 ),

CN = BN

Thus, CN / BN = 1 --- ( 3 )

From ( 2 ),

AM = DM

Thus, AM / DM = 1 --- ( 4 )

From ( 3 ) and ( 4 ),

AM / DM = CN / BN