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$ \triangle PQR \ $ is an isosceles triangle where $PQ = PR. $ $X$ is a point on the circumcircle of $ΔPQR $, such that it lies on the arc $QR$. The normal drawn from the point $P$ on $XR$ intersects $XR$ at point $Y.$ Prove that $$QX + YR = XY.$$

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Reflect the $R$ across line $PY$ in to $Z$. So $Z\in XR$ and $YZ = YR$ and $$PZ = PR = PQ$$

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Sp we have to prove that $QX = XZ$. Since $$\angle XZP = \pi-\angle RZP = \pi-\angle XRP = \angle XQP$$ and $$\angle QXP = \angle QRP = \angle RQP = \angle ZXP$$ we see that triangle $XQP$ is congruent to triangle $XZP$ (s.a.s.) and we are done.

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  • $\begingroup$ I think the reason of congruence is AAS instead. $\endgroup$
    – Mick
    Dec 21 '18 at 2:24
  • $\begingroup$ Could be also ASA @Mick $\endgroup$
    – Aqua
    Dec 21 '18 at 8:06
  • $\begingroup$ I mean at least not (s. a. s.). $\endgroup$
    – Mick
    Dec 21 '18 at 17:32
  • $\begingroup$ And why not @Mick $\endgroup$
    – Aqua
    Dec 21 '18 at 17:59
  • $\begingroup$ In your proof, you have mentioned 2 pairs of equal angles (and a common side). That means the reason to be quoted is either AAS or ASA. $\endgroup$
    – Mick
    Dec 22 '18 at 8:51

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