# In ∆ABC, AB=AC, $\angle A = 80°$, find angles of ∆AXY

So the question is,

In a triangle $$ABC$$, $$AB=AC$$, $$A=80°$$ and $$S$$ is circumcentre. Bisectors of angles $$ACS$$ and $$ABS$$ meet $$BS$$ and $$CS$$ at $$X$$ and $$Y$$ respectively. find angles of $$∆AXY$$ I found all the angles till the very last $$∆AXY$$ is left but then I'm not able to find any way of calculating any angle further. I found out all the similar triangles but it's not helping. I'm attaching my worked out image below:

My workings: $$\angle BAC$$=80°, $$\angle BSC=160°$$, $$BS=BC$$, Thus, $$\angle SBC=SCB=10°$$ $$\angle ABS=ACS=40°$$, $$∆BSX$$is congruent to $$∆CSY$$ and both of them are isosceles with the base angles of $$20°$$

$$P$$ is the point of intersection of $$AS$$ and $$XY$$. $$∆XSP$$ and $$∆YSP$$ are congruent, thus angle $$XPS=90°= \angle APX$$ $$∆APX$$is congruent to $$∆APY$$. I don't know now what to do. Some insights would be helpful.

• I am not very sure how you proved that $BSC = 160$. Could you add details please? Apr 10, 2021 at 10:33
• S is the circumcentre, and A also lies on the circle. So inscribed angle theorem Apr 10, 2021 at 10:34
• That's fair. Thank you Apr 10, 2021 at 10:36
• The figure is completely symmetric. Look for congruent triangles. It turns out to be an equilateral triangle. Apr 10, 2021 at 10:50
• I know that the two triangles are congruent but this just means that $\angle AXP = \angle AYP$ and $\angle YAP = \angle XAP$...which just gives $\angle AXP + \angle YAP = 90°$ Apr 10, 2021 at 11:28

The figure is symmetric so knowing $$\angle BAX$$ will do the work.

The problem can be differently worded as follows :

In $$\triangle ABC$$, $$AB=AC$$ and $$\angle A=80^{\circ}$$. $$X$$ is a point inside the triangle for which $$\angle XBC=30^{\circ}$$ and $$\angle XCB=20^{\circ}$$. Find $$\angle BAX$$.

Let the circumcentre of $$\triangle BXC$$ be $$O$$.

Observe that, $$\triangle BOC\cong \triangle BAC$$ and $$\triangle OXC$$ is equilateral from angle chasing. Hence, $$\triangle XAC$$ is isosceles and thus $$\angle BAX=10^{\circ}$$.

• How is $\triangle BOC\cong \triangle BAC$ ? The angles are all different as per my calculations in the diagram Apr 10, 2021 at 11:49
• And Point X, O(the circumcentre) and C are supposed to be collinear then how can ∆OXC be a triangle? (In my diagram, S is the circumcentre instead of O) Apr 10, 2021 at 11:57
• @Ruchi $\triangle BOC\cong \triangle BAC$ because they are both isosceles with $AB=AC$ and $OB=OC$, share the common side $BC$ and $\angle BOC=80^{\circ}=\angle BAC$. Apr 10, 2021 at 12:37
• @Ruchi The points $X$, $O$ and $C$ are not collinear. $O$ is the circumcentre of $\triangle BXC$ Apr 10, 2021 at 12:39