0
$\begingroup$

Let $ABC$ be a triangle with $AC > AB$. The point $X$ lies on the side $BA$ extended through $A$, and the point $Y$ lies on the side $CA$ in such a way that $BX = CA$ and $CY = BA$. The line $XY$ meets the perpendicular bisector of side $BC$ at $P$. Show that $\angle BPC + \angle BAC = 180^{\circ}$.

$\endgroup$
1
$\begingroup$

Let $l$ denote the angle bisector of $\angle{BAC}$ and $m$ the parallel to $l$ passing through the midpoint $M$ of $BC$. It suffices to prove that $XY$ is reflection of $l$ over $m$. Indeed, since midpoint of arc $BC$ of the circumcircle of $ABC$ which doesn't contain $A$ lies on $l$ and $M$ lies on $m$, the reflection of the first one with respect to the second one lies on $XY$ which happens to be a point $P$. Since the midpoint of arc satisfies the equality of angles in question and reflection with respect to $M$ preserves this property we get the result. Now, we will prove fact mentioned at the beginning it in three different ways:
1) If we denote the intersection of $l$ and $BC$ by $L$ as well as the intersection of $XY$ and $BC$ by $Z$, by the angle bisector and Thales theorem (as previously mentioned $AL || XY$) we have: $$\frac{CZ}{CY}=\frac{CL}{CA}=\frac{BL}{BA}$$ and since $CY=BA$ we have $BL=CZ$ i.e. $Z$ and $L$ are symmetric with respect to $M$ which yields the result.
2) If we consider glide reflection with respect to $m$ such that the image of $AB$ is $AC$ we notice that $X$ goes to $A$ (because $B$ goes to $C$ in this reflection as the midpoint of $BC$ stays fixed). We can consider analogous reflection taking $AC$ to $AB$. This proves that the distances of $A$, $X$ and $Y$ to $m$ are the same which finishes the proof.
3) One can also consider $X'$ and $Y'$ to be the midpoints of $AX$ and $AY$ and make use of the fact that $AX'=\frac{1}{2}AX=\frac{BX-AB}{2}=\frac{b-c}{2}$ and similarly for $AY'$ by applying Menelaus theorem to $X'$, $Y'$ and $M$ to see that they are collinear.

$\endgroup$
0
$\begingroup$

HINT.

The line through $A$ parallel to $XY$ is the bisector of angle $\angle BAC$. It meets again the circle through $ABC$ at the midpoint of arc $BC$, i.e. on the perpendicular bisector of $BC$.

$\endgroup$
  • $\begingroup$ Please give full solution $\endgroup$ – user671269 May 6 '19 at 15:23
0
$\begingroup$

A few things to note: $PB=PC$, since $P$ is on the perp. bisector of $BC$. Also, we have $AC=BX$ and $CY=AB$ so $AY=AX$. Also, do you see that the claim is proven if you can prove that $\angle BPC=\angle XAY$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.