Length relation between middle point, incircle touching point, angle bisector foot and altitude foot. $DE^2=DF\cdot DP$ I have the following exercise.
In the $\triangle ABC$, $D$ is the middle point of $AB$, $P$ is the foot of the altitude $CP$ on $AB$, the angle bisector $CF$ intersecting $AB$ at $F$, $E$ is the touching point of the incircle on $AB$. Prove that
$$
DE^2=DF\cdot DP
$$

I can prove it by computing all segments:
$DB=\frac{c}{2}$, $BP=a\cdot\frac{b^2-a^2-c^2}{2ac}=\frac{b^2-a^2-c^2}{2c}$, $BF=\frac{ac}{a+b}$, so $DF=DB-BF=\frac{c(b-a)}{2(b+a)}$, $DP=DB+BP=\frac{b^2-a^2}{2c}$, $DE=\frac{c}{2}-(p-b)=\frac{b-a}{2}$. Hence,
$$
DE^2=DF\cdot DP
$$
But I want to see how to prove this by a geometrical way. Let $DG$ be the other tangent line from $D$ to the incircle. So $DG=DE$. From this resulting identity, I need to show $\triangle DGP\sim\triangle DFG$. I only know that $\angle DGF=\angle GTS$. So it suffices if I show $ST\parallel DP$. But I could not prove $ST\parallel DP$. Do I miss some properties?
 A: Another approach. Let add to the picture $c_1$ — circumcircle of ABC. Let K is middle of arc AB. Then from similarity of triangles KDF, IEF, CPF follows that we need to prove $KI^2=KF\cdot KC$.
Let consider triangle BKI. $\angle BKI=\angle BAC=\alpha$. $\angle KBI=\angle KBA+\angle ABI=\angle KCA+\angle ABI=\gamma/2+\beta/2$. $\angle KIB=\pi-\angle BKI-\angle KBI=\pi-\alpha-\beta/2-\gamma/2=\gamma/2+\beta/2=\angle KBI$. Then $KI=KB=KA$. Circle $c_2$ is not used in proof, shown only for demonstration.
Let add $c_3$ — circumference of AFC. $\angle FAK=\angle FCB=\angle FCA$, then KA is tangent to $c_3$. Then $KI^2=KA^2=KF\cdot KC$, q.e.d.

A: I give another proof. I got this idea from @Ivan Kaznacheyeu's proof and other exercises I did recently.

*

*Let $G$ be the middle point of $BC$, connect $DG$ intersecting $CF$ at $S$. Connect $BS$,$ES$,$PS$. $DG\parallel AC$.

*$\angle GSC=\angle DSF=\angle ACF=\angle SCG=\frac{\angle ACB}{2}$, so $SG=CG=\frac12BC$. $\angle BSC=90^\circ$.

*$CP\bot AP$,so $B,S,C,P$ are cyclic. $\angle DPS=\angle BCS=\frac{\angle ACB}{2}$.

*$\angle DSF=\angle DPS$, so $\triangle DSF\sim \triangle DPS$, $DS^2=DF\cdot DP$.


Next, I prove that $DS=DE$ with another picture. I remove unnecessary points and segments.


*$I$ is the incenter, connect $IE$, $IB$. $\angle BSI=\angle BEI=90^\circ$, so $B,I,S,E$ are cyclic. $\angle IES=\angle IBS$.

*$\angle SFB=A+\frac{C}{2}$, $\angle SBF=90^\circ-\angle SFB=\frac{B-A}{2}$.

*$\angle EBI=\frac{B}{2}$, so $\angle IBS=\angle IBE-\angle EBS=\frac{A}{2}$.

*$\angle IES=\frac{A}{2}$, $\angle SED=90^\circ-\angle IES=90^\circ-\frac{A}{2}$.

*In $\triangle DES$, $\angle EDS=A$, $\angle SED=90^\circ-\frac{A}{2}$, so $\angle ESD=90^\circ-\frac{A}{2}$.

So $DE=DS$. The proof is done.

