Question

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!

Answer 1

see explanation.

Explanation:

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^@`