Question

In a triangleABC, AB=AC. Suppose the bisector of angleACBmeets AB at M. Let AM+MC=BC. Prove angleBAC= ${100}^{\circ }$

Answer 1

Construct MD with D on BC such that MD = BD

Let MD =BD.
$\Delta MDBisisocelers.\angle BMD=\angle MBD=\alpha \Rightarrow \angle AMD={180}^{\circ }-\alpha .$
Consider Quadrilatral AMDC.
$\angle AMD={180}^{\circ }-\alpha$ & $\angle ACD=\alpha$
$\therefore \angle AMD+\angle ACD={180}^{\circ }$. Since they are opposite
$\angle BAC={180}^{\circ }-2\alpha$
Join $AD.\angle MAD=\angle MCD=\frac{\alpha }{2}$ (angle is same segment)
& $\angle ADM=\angle ACM=\frac{\alpha }{2}$
$\therefore \angle MAD=\angle ADM\Rightarrow \Delta ADM$ is isocesess.
Join $AD.\angle MAD=\angle MCD=\frac{\alpha }{2}$ (angle is same segrment)
& $\angle ADM=\angle ACM=\frac{\alpha }{2}$
$\therefore \angle MAD=\angle ADM\Rightarrow \Delta ADM$ is isocesess.
$\Rightarrow AM=MD.$
But RD =BD $\Rightarrow$ AM =BD.
Now BC =AM +MC (Given)
=BD +MC
=BD +DC
$\Rightarrow CM=DC\Rightarrow \Delta CMDisisoceler.\therefore \angle CDM=\angle CMD=2\alpha \therefore \angle CMD=\alpha +2\alpha =3\alpha =\angle BAC+\frac{\alpha }{2}={120}^{\circ }-2\alpha +\frac{\alpha }{2}5\alpha -\frac{\alpha }{2}={180}^{\circ }\frac{a\alpha }{2}={180}^{\circ }\Rightarrow \alpha ={40}^{\circ }\angle BAC={180}^{\circ }-2\left(40\right)={180}^{\circ }-Proved$