Let #angleABC and angleACB " be " 2beta, => angleACM=angleMCB=beta#
Mark point #D# on #BC#, such that #CD=CM#.
Draw a line #ME#, parallel to #BC#, as shown in the figure,
#=> angleEMC=angleMCB=beta#,
and #angleAME=angleAEM=2beta#
let #CM=x, AM=y and MB=z#,
#=> CD=x, AE=y and EC=z#,
In #DeltaEMC#, as #angleEMC=angleECM=beta, => DeltaEMC# is isosceles,
#=> EM=EC=z#
As #BC=AM+MC=BD+CD, => y+x=BD+x, => BD=y=AM#
Now, as #BM=ME=z, BD=MA=y#, and included angle #angleDBM=angleAME=2beta#,
#=> DeltaDBM and DeltaAME# are congruent,
#=> angleDMB=angleAEM=2beta#
Now, #angleCDM=angleDBM+angleDMB=2beta+2beta=4beta#
As #CD=CM, => angleCMD=angleCDM=4beta#,
In #DeltaCDM, 4beta+4beta+beta=9beta=180^@#,
#=> beta=20^@#
Hence, #angleBAC=180-2beta-2beta=180-4beta=180-(4xx20)=100^@#
