In the diagram below of isosceles triangle ABC, AB is congruent to CB and angle bisectors AD, BF and CE are drawn and intersect at x.If the measure of angle BAC=50°, find the measure of AXC
Accepted Solution
A:
Answer:[tex]130^{\circ}[/tex]Step-by-step explanation:We know that in an triangle , angles opposite to equal sides are equal .Given : [tex]AB=CB[/tex]So, [tex]\angle A=\angle C[/tex] . We get [tex]\angle C=\angle A=50^{\circ}[/tex]Since AD bisects [tex]\angle A[/tex] , [tex]\angle XAC=\frac{1}{2}\angle A=\frac{1}{2}(50^{\circ})=25^{\circ}[/tex]Also, as CE bisects [tex]\angle C[/tex] , [tex]\angle XCA=\frac{1}{2}\angle C=\frac{1}{2}(50^{\circ})=25^{\circ}[/tex]We will use angle sum property which states that in [tex]\Delta XAC[/tex] ,[tex]\angle XAC+\angle XCA+\angle AXC=180^{\circ}[/tex][tex]25^{\circ}+25^{\circ}+\angle AXC=180^{\circ}\\50^{\circ}+\angle AXC=180^{\circ}\\\angle AXC=180^{\circ}-50^{\circ}\\=130^{\circ}[/tex]