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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:

enter image description here

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.

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  • $\begingroup$ I am not very sure how you proved that $BSC = 160$. Could you add details please? $\endgroup$
    – Gabrielek
    Apr 10, 2021 at 10:33
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    $\begingroup$ S is the circumcentre, and A also lies on the circle. So inscribed angle theorem $\endgroup$
    – Ruchi
    Apr 10, 2021 at 10:34
  • $\begingroup$ That's fair. Thank you $\endgroup$
    – Gabrielek
    Apr 10, 2021 at 10:36
  • $\begingroup$ The figure is completely symmetric. Look for congruent triangles. It turns out to be an equilateral triangle. $\endgroup$
    – krazy-8
    Apr 10, 2021 at 10:50
  • $\begingroup$ 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 $\angle AXP + \angle YAP = 90°$ $\endgroup$
    – Ruchi
    Apr 10, 2021 at 11:28

1 Answer 1

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Triangle ABC

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 $\angle BAX=10^{\circ}$.

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  • $\begingroup$ How is $\triangle BOC\cong \triangle BAC$ ? The angles are all different as per my calculations in the diagram $\endgroup$
    – Ruchi
    Apr 10, 2021 at 11:49
  • $\begingroup$ 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) $\endgroup$
    – Ruchi
    Apr 10, 2021 at 11:57
  • $\begingroup$ @Ruchi $\triangle BOC\cong \triangle BAC$ because they are both isosceles with $AB=AC$ and $OB=OC$, share the common side $BC$ and $\angle BOC=80^{\circ}=\angle BAC$. $\endgroup$
    – Limestone
    Apr 10, 2021 at 12:37
  • $\begingroup$ @Ruchi The points $X$, $O$ and $C$ are not collinear. $O$ is the circumcentre of $\triangle BXC$ $\endgroup$
    – Limestone
    Apr 10, 2021 at 12:39

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