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Let $ABC$ be an isosceles triangle with $AB=AC$. A point $P$ on the same side as $A$ with respect to $BC$ satisfies $\angle BPC=90^\circ+\frac{1}{2}\angle BAC$. Three line segments $MN$, $PK$, and $PL$ are drawn through the point $P$ in such a way that $MN\parallel BC$, $PK\parallel AB$, and $PL\parallel AC$. The straight lines $MK$ and $NL$ meet at a point $D$. Prove that $ABCD$ is a cyclic quadrilateral.

you can also use this shape.

my attempt:

I draw the circle which passes from $B, P, C$ and I tried to continue to a solution but I wasn't successful.it can also help you: if you can prove $PKDC$ is cyclic quadrilateral which can be proved by showing $NPDC$ is cyclic quadrilateral the problem will be solved.

note: please share your ideas and your attempts in comments or even answer if your solution is not complete.thanks!

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We'll try to prove that $\triangle PLC \sim \triangle BKP$. Obviously we have that $\angle PLC = \angle CKP = 180 - \angle C$. Now:

$$\angle LPC + \angle PCL = \angle C$$ $$\angle PBC + \angle PCB = 180 - \angle BPC = 180 - 90 - \frac 12 \angle A = \angle C$$

From this two equalities, as $\angle PCB = \angle PCL$ we have that $\angle LPC = \angle PBC$, which proves the similarity of the triangles. Now using this we'll prove that $\triangle MPK \sim \triangle LPN$. As $\angle MPK = \angle NPL = \angle C$ it's enough to prove that $\frac{MP}{PK} = \frac{PL}{PN}$. This is true as:

$$\frac{MP}{PK} = \frac{BK}{PK} = \frac{PL}{LC} = \frac{PL}{PN}$$

where the second equality follows from the fact we initially proved. Now we get:

$$\angle NDM = 180 - \angle DNM - \angle DMN = 180 - \angle DNM - \angle NLP = \angle LPN = \angle C$$

So now $\angle KDN = \angle C = \angle DCN$, so $KDCN$ is cyclic. Also $KCNP$ is trivially cyclic, so all the $5$ points lie on the same circle. Similarly $BMPLD$ is cyclic so now:

$$\angle BDC = \angle BDL + \angle CDK - \angle KDL = 180 - \angle BPL + 180 - \angle CPK - \angle C = $$ $$360 - \angle BPC - \angle KPL - \angle C = 270 - \frac 32 \angle A - \angle C = 180 - \frac 32 \angle A + \frac 12 \angle A = 180 - \angle A$$

Which proves that $ABCD$ is cyclic.

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  • $\begingroup$ it was excellent! thanks! $\endgroup$ – math enthusiastic Dec 30 '17 at 11:19
  • $\begingroup$ I think it should be $ BMPLD$, not $AMPLD$ when you mentioned similarly... $\endgroup$ – math enthusiastic Dec 30 '17 at 11:29
  • $\begingroup$ @f.jalili Just a typo $\endgroup$ – Stefan4024 Dec 30 '17 at 12:34

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