Let angleABC and angleACB " be " 2beta, => angleACM=angleMCB=beta∠ABCand∠ACB be 2β,⇒∠ACM=∠MCB=β
Mark point DD on BCBC, such that CD=CMCD=CM.
Draw a line MEME, parallel to BCBC, as shown in the figure,
=> angleEMC=angleMCB=beta⇒∠EMC=∠MCB=β,
and angleAME=angleAEM=2beta∠AME=∠AEM=2β
let CM=x, AM=y and MB=zCM=x,AM=yandMB=z,
=> CD=x, AE=y and EC=z⇒CD=x,AE=yandEC=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^@