The angle bisector of $\angle ABC$ of triangle $ABC$ ($AB>BC$) cuts the circumcircle of that triangle in $K$. The foot of the perpendicular from $K$ to $AB$ is $N$, and $P$ is the midpoint of $BN$. The line through $P$ parallel to $BC$ cuts line $BK$ in $T$. Prove that the line $NT$ passes through the midpoint of $AC$.
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Tags: geometry, angle bisector
SCLT
10.08.2020 18:47
Steve12345 wrote: The angle bisector of $\angle ABC$ of triangle $ABC$ ($AB>BC$) cuts the circumcircle of that triangle in $K$. The foot of the perpendicular from $K$ to $BC$ is $N$, and $P$ is the midpoint of $BN$. The line through $P$ parallel to $BC$ cuts line $BK$ in $T$. Prove that the line $NT$ passes through the midpoint of $AC$. You say that $N$ is on $BC$, which means that $P$ is also on $BC$. So what is the line through $P$ parallel to $BC$? Is there a typo somewhere?
Steve12345
10.08.2020 18:50
SCLT wrote: You say that $N$ is on $BC$, which means that $P$ is also on $BC$. So what is the line through $P$ parallel to $BC$? Is there a typo somewhere? Edited.
xoxo.
10.08.2020 19:28
Hopefully this is right!
Let $E$ be the midpoint of $AC$, $M$ be the foot of $K$ onto $BC$, $T'=BK\cap MN$. Notice that $NE$ is just the Simson line of $\triangle ABC$ with respect to $K$, so $M$ also lies on this line.
Claim: $\triangle BT'N\cong\triangle BT'M$
Proof: Since $BK$ is the angle bisector of $\angle ABC$, $\angle T'BN=\angle T'BM$. Also, using the fact that $CEKM$ is cyclic, $\angle BMN=\angle CME=\angle CKE=90-\angle ACK=90-\angle ABK$ and similarly $\angle BNM=90-\angle CBK$, but since $\angle ABK=\angle CBK$ we have $\angle BMN=\angle BNM$. Now, since the triangles share common side $BT'$, they must be congruent by AAS as desired.
Note that $NT'=T'M$, at which point $PT'$ is the midline of $\triangle BNM$ and thus must be parallel to $BC$ as desired.
Steve12345
10.08.2020 20:25
Faster: Let $S$ be the midpoint of $AC$. Notice that $SK \perp AC$. Let $M$ be the other foot from $K$ to $BC$. $(M,S,N)$ are collinear by Simsons line. By easy angle chasing we get $NT \perp BK$ and thus since $MN \perp BK$ we get $(N,S,T)$ are collinear too, hence proving the claim.
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rafigamath
25.05.2023 15:02
Let $\angle{CBT}={\alpha}$.Since $\angle{ABT}=\angle{TBC}$ and $ TP\parallel BC \implies PT=BP=PN$.So $\angle{BTN}={90},\implies \angle{KNT}=\angle{NBT}={\alpha}$ . Now let $NT \cap AC=X$. If we prove that $KX\perp AC$ $\implies ANXK$ cyclic we're done. And this is obvious since $\angle{ABT}=\angle{KAX}, \angle{ABT}=\angle{XNK}$.
wizixez
15.12.2024 11:03
Set $(ABC) \in \mathbb{S}^1$ $a=x^2$ , $b=y^2$ , $c=z^2$ $k=-xz$ , $2n=x^2+y^2-xz-x^2z^2k'$ $2p=b+n$ and finally $t=bc/p$ We want to prove that $NT \cap AC=M$ and $m=\frac{nt(a+c)-ac(n+t)}{nt-ac}=\frac{a+c}{2}$ Which is true if you do the "calculations"