Question

Please answer its urgent ​

Answer 1

∆AMO~∆BMO(From SAS)

then

OAM=OBM

AOM=BOM

OMA=OMB

Since, OMA and OMB is on line AB

then

OMA+OMB=180

2 OMA =180

OMA. = 90

OMB. = 90

then

OM perpendicular to AB

Answer 2

soln-given,

Toprove-OM is perpendicular to AB

proof-in triange AMO &BMO

Aom=Bom

OmA=OmB=90dproved

MAo=MBo

Therefore,Ao=BO

<oMA=<oMB(=90)

0M=OM(common side)

therefore,∆AMo~∆BMo(SAs)

since,∆oMA & ∆oNB is on line AB

oMA+OMB=180°

oMA+oMA=180°(from i)

2 oMA=180°

OMA=180°/2

oMA=90°

therefore,OM is perpendicular to AB.

Hence pproved