An appearance of harmonic division

Consider a semicircle $(AB)$ and a circle, $w$, tangent internally at $T$. Let $N$ be the point of tangency of $w$ and the diameter, $AB$. If $O$ is the center of $(AB)$ and $C$ is the midpoint of arc $AB$, let $F$ be the intersection of a parallel line to $CT$ passing through $D$ with $CO$. Denote $P$ the intersection of $CN$ with $DF$. Prove that $DP=FP$. (Carlos Hugo Olivera Diaz)

Proof. It is a well-known fact that $O$, $D$ and $T$ are collinear, so we have that $\triangle{OFD}\sim{\triangle{OCT}}$, as a consequence, $DT=CF=DN=r$, where $r$ is the radius of $w$. Since $CO\parallel{DN}$, $\angle{FCP}=\angle{DNP}$. Also, $\angle{FPC}=\angle{NPD}$, hence $\triangle{CFP}\cong{\triangle{DNP}}$ and the proof is complete. 

$\square$

Remark 1. If $CN$ meet $OT$ at $H$, then, $(O, D; H, T)=-1$.

Proof 1. If we project from $C$ onto the line $OT$, we have $-1=(F, D; P, P_\infty)=(O, D; H, T)$. 

$\square$

Proof 2. Let $N'$ be the second intersection of $ND$ with $w$. Then, from the first lemma in the Archimedes's Book of Lemmas, we know that $C$, $N'$ and $T$ are collinear. As $D$ is the midpoint of $NN'$ and $NN'$ is parallel to $CO$, if we project from $C$ onto the line $OT$, $-1=(N', N; D, P_{\infty})=(T, H; D, O)$.

$\square$

Notice that the original problem can also be proven using harmonic division. Indeed, projecting from $C$ onto the line $DF$, we have $-1=(N', N; D, P_{\infty})=(P_{\infty}, P, D, F)$, which means $P$ is the midpoint of $DF$.

External version.  Here $D'$ is the reflection of $D$ around $N$. Let $H$ be the intersection of $OD$ with $CD'$, then, $(O, T; H, D)=-1$.

Proof. As mentioned, it is known that $O$, $T$ and $D$ are collinear. Also, because of the external version of Archimedes's lemma, $C$, $T$ and $N$ are collinear. As $CO\parallel{DD'}$, if we project from $C$ onto the line $OD$, $-1=(D, D'; N, P_{\infty})=(D, H; T, O)$.

$\square$

Remark 2. In this case, let $H$ be the intersection of $CT$ with $AB$. Then, $(A, B; N, H)=-1$. (See problems 2 and 3 for applications)

Proof. From Archimedes's lemma we know that $C'$, $N$ and $T$ are collinear. The quadrilateral $ACBC'$ is harmonic, a square specifically, so if we project from $T$ onto the line $AB$ we have $-1=(A, B; C', C)=(A, B; N, H)$. 

$\square$

In general, if $N$ and $H$ are the intersections of $C'T$ with $AB$ and $CT$ with $AB$, respectively, $AB$ could be any chord perpendicular to $CC'$ and $T$ be any point on the circle $(ACBC')$ and the result still holds. This is because $ACBC'$ is always a harmonic quadrilateral (e.g. a square or a kite).

Applications.

Problem 1. Let $w$ and $v$, be two circles tangent internally at $T$. Let $N$ be the point of tangency of the smaller circle, $v$, and the diameter of $w$, $AB$. If $C$ is the midpoint of arc $AB$ (above $AB$), let's $C'$ be its antipode. Also, let $N'$ be the antipode of $N$. If $O$ is the center of circle $w$, prove that $CN$, $C'N'$ and $OT$ are concurrent at $P$.

Proof. Let's supose $CN$ meets $OT$ at $P$ and let $D$ be the center of $v$. From the remark discussed previously, we know that $(O, D; P, T)=-1$. We also know from Archimedes's lemma that $C$, $N'$ and $T$ are collinear. So if we project from $N'$ onto the line $CO$ we have that $T$ goes to $C$; $O$ remains the same and $D$ goes to the point at infinity, hence, $N'$, $P$ and  $C'$ must be collinear and the proof is complete.

The following is problem 310 in Gogeometry. I have provided two more proofs here (in Spanish).

Problem 2. Let $w$ and $v$, be two circles tangent internally at $T$. Let $N$ be the point of tangency of the smaller circle, $v$, and the diameter of $w$, $AB$. Prove that $\angle{ATN}=45^\circ$.

Proof. Let $C$ be the midpoint of arc $AB$ (containing $T$) and let $C'$ be its antipode. Because of Archimedes's lemma $C'$, $N$ and $T$ are collinear. If $H$ is the intersection of $CT$ with $AB$, from remark 2 we know that $(A, B; N, H)=-1$. Also, notice that $\angle{CTC'}=\angle{NTH}=90^\circ$, since $CC'$ is a diameter. Therefore, $NT$ is an angle bisector of $\angle{ATB}=90^\circ$ and the proof is complete.

Problem 3. Let $D$ be the center of a circle inscribed in a semicircle $(AB)$. Let $T$ and $N$ be the points where the circle touches $(AB)$ and $AB$, respectively. Call $C$ the midpoint of arc $AB$. Prove that $AC$, $ND$ and $BT$ are concurrent at a common point. 

Proof. Let $H$ be the intersection of $CT$ and $AB$ and $P$ the intersection of $AC$ and $BT$. From the remark 2 we know that $(A,B;N,H)=-1$. This implies that $AT$, $BC$ and $NP$ are concurrent at a common point. Since $\angle{ACB}=\angle{BTA}=90^\circ$, the segment $NP$ must be an altitude of $\triangle{ABP}$. As $\angle{BND}=\angle{BNP}=90^\circ$, we conclude that $AC$, $ND$ and $BT$ are concurrent at a common point and the proof is complete. 

Related material:

Harmonic Divisions (Lemmas)