Let ABC be a triangle with AB=AC. The bisector of angle ACB meet AB at M. Suppose AM+MC=BC. Show that angle BAC = 100 degrees? Thank you!

1 Answer
CW
Jan 14, 2018

see explanation.

Explanation:

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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^@

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