Prove that $DG+HE=GH$ for the given figure.

Question

What I could gather:

$$\measuredangle ODB=\measuredangle OFC=\measuredangle OEC =90 \text{ degrees}$$ $$OD=OF=OE=r$$ $$\measuredangle ODE=\measuredangle OED$$ $$\measuredangle DOE=2\measuredangle DME$$

Answer 1

This is a pure trigonometry solution.

Let $|BC|=a$, $|AC|=b$, $|AB|=c$, $\angle BAC=\alpha$, radius of excircle $|O_aM|=r_a$, $S$ is the area and $\rho=\tfrac12(a+b+c)$ is a semiperimeter of $\triangle ABC$.

Let's prove a slightly stronger statement than $|DG|+|EH|=|GH|$, namely that

\begin{align} |DG|&=|LH|,\quad |GL|=|EH| . \end{align}

\begin{align} |AM|&=r_a\,\frac{1+\sin\tfrac\alpha2}{\sin\tfrac\alpha2} ,\quad |AL|=r_a\,\frac{\cos^2\tfrac\alpha2}{\sin\tfrac\alpha2} ,\\ |AK|&=c\,\cos\tfrac\alpha2 ,\quad |AN|=b\,\cos\tfrac\alpha2 ,\quad |LM|=r_a\,(1+\sin\tfrac\alpha2) ,\\ |KM|&=r_a\,\frac{1+\sin\tfrac\alpha2}{\sin\tfrac\alpha2} -c\,\cos\tfrac\alpha2 ,\\ |NM|&=r_a\,\frac{1+\sin\tfrac\alpha2}{\sin\tfrac\alpha2} -b\,\cos\tfrac\alpha2 ,\\ |BK|&=c\,\sin\tfrac\alpha2 ,\quad |CN|=b\,\sin\tfrac\alpha2 ,\\ |DL|&=|LE|=r_a\,\cos\tfrac\alpha2 ,\\ \end{align}

\begin{align} |GL|&=\frac{|LM|\,|BK|}{|KM|} =\frac{c\,r_a\,(1+\sin\tfrac\alpha2)\,\sin\tfrac\alpha2} {r_a\,\frac{1+\sin\tfrac\alpha2}{\sin\tfrac\alpha2} -c\,\cos\tfrac\alpha2} ,\\ |HL|&=\frac{|LM|\,|CN|}{|NM|} = \frac{b\,r_a\,(1+\sin\tfrac\alpha2)\,\sin\tfrac\alpha2} {r_a\,\frac{1+\sin\tfrac\alpha2}{\sin\tfrac\alpha2} -b\,\cos\tfrac\alpha2} ,\\ |EH|&=r_a\,\cos\tfrac\alpha2 - \frac{b\,r_a\,(1+\sin\tfrac\alpha2)\,\sin\tfrac\alpha2} {r_a\,\frac{1+\sin\tfrac\alpha2}{\sin\tfrac\alpha2} -b\,\cos\tfrac\alpha2} \end{align}

Now we just need to prove that $|GL|=|EH|$ is an identity:

\begin{align} \frac{c\,(1+\sin\tfrac\alpha2)\,\sin\tfrac\alpha2} {r_a\,\frac{1+\sin\tfrac\alpha2}{\sin\tfrac\alpha2} -c\,\cos\tfrac\alpha2} = \cos\tfrac\alpha2 - \frac{b\,(1+\sin\tfrac\alpha2)\,\sin\tfrac\alpha2} {r_a\,\frac{1+\sin\tfrac\alpha2}{\sin\tfrac\alpha2} -b\,\cos\tfrac\alpha2} \tag{1}\label{1},\\ \frac{c\,(1+\sin\tfrac\alpha2)\,\sin\tfrac\alpha2} {r_a\,\frac{1+\sin\tfrac\alpha2}{\sin\tfrac\alpha2} -c\,\frac{1-\sin^2\tfrac\alpha2}{\cos\tfrac\alpha2}} = \cos\tfrac\alpha2 - \frac{b\,(1+\sin\tfrac\alpha2)\,\sin\tfrac\alpha2} {r_a\,\frac{1+\sin\tfrac\alpha2}{\sin\tfrac\alpha2} -b\,\frac{1-\sin^2\tfrac\alpha2}{\cos\tfrac\alpha2}} \tag{2}\label{2},\\ \frac{c\,\sin\tfrac\alpha2} {\frac{r_a}{\sin\tfrac\alpha2} -c\,\frac{1-\sin\tfrac\alpha2}{\cos\tfrac\alpha2}} = \cos\tfrac\alpha2 - \frac{b\,\sin\tfrac\alpha2} {\frac{r_a}{\sin\tfrac\alpha2} -b\,\frac{1-\sin\tfrac\alpha2}{\cos\tfrac\alpha2}} \tag{3}\label{3},\\ \frac{c\,\sin^2\tfrac\alpha2} {r_a\,\cos\tfrac\alpha2 -c\,(1-\sin\tfrac\alpha2)\,\sin\tfrac\alpha2} = 1 - \frac{b\,\sin^2\tfrac\alpha2} {r_a\,\cos\tfrac\alpha2 -b\,(1-\sin\tfrac\alpha2)\,\sin\tfrac\alpha2} \tag{4}\label{4},\\ \frac{\sin^2\tfrac\alpha2} {\frac{r_a}{c}\,\cos\tfrac\alpha2 -(1-\sin\tfrac\alpha2)\,\sin\tfrac\alpha2} + \frac{\sin^2\tfrac\alpha2} {\frac{r_a}{b}\,\cos\tfrac\alpha2 -(1-\sin\tfrac\alpha2)\,\sin\tfrac\alpha2} = 1 \tag{5}\label{5},\\ \frac{1} {\frac{r_a}{c}\,\cos\tfrac\alpha2 -(1-\sin\tfrac\alpha2)\,\sin\tfrac\alpha2} + \frac{1} {\frac{r_a}{b}\,\cos\tfrac\alpha2 -(1-\sin\tfrac\alpha2)\,\sin\tfrac\alpha2} = \frac1{\sin^2\tfrac\alpha2} \tag{6}\label{6},\\ . \end{align}

This boils down to

\begin{align} \sin^4\tfrac\alpha2-\sin^2\tfrac\alpha2 - \frac{r_a^2}{bc}\,\cos^2\tfrac\alpha2 +\sin\tfrac\alpha2\,\cos\tfrac\alpha2 \left(\frac{r_a}{b}+\frac{r_a}{c}\right) &=0 \tag{7}\label{7},\\ bc\,\sin^2\tfrac\alpha2\,(\sin^2\tfrac\alpha2-1) - r_a^2\,\cos^2\tfrac\alpha2 +\sin\tfrac\alpha2\,\cos\tfrac\alpha2 (r_a\,b+r_a\,c) &=0 \tag{8}\label{8},\\ -bc\,\sin^2\tfrac\alpha2\,\cos\tfrac\alpha2 - r_a^2\,\cos\tfrac\alpha2 +\sin\tfrac\alpha2\, (r_a\,b+r_a\,c) &=0 \tag{9}\label{9},\\ r_a\,(b+c)-bc\,\sin\tfrac\alpha2\,\cos\tfrac\alpha2-r_a^2\,\cot\tfrac\alpha2 &=0 \tag{10}\label{10},\\ r_a\,(b+c)-S-r_a^2\,\cot\tfrac\alpha2 &=0 \tag{11}\label{11},\\ r_a\,(b+c-r_a\,\cot\tfrac\alpha2) &=S \tag{12}\label{12},\\ 2\,(b+c-r_a\,\cot\tfrac\alpha2) &=-a+b+c \tag{13}\label{13},\\ \tfrac12(a+b+c) &= r_a\,\cot\tfrac\alpha2 \tag{14}\label{14},\\ \tan\tfrac\alpha2 &\equiv \frac{S}{\rho\,(\rho-a)} \tag{15}\label{15} , \end{align}

which is a well-known identity.

Answer 2

This is inversion proof (not projective!) with respect to circle which center is $M$ and $r= MD = ME$.

Let $FM$ cuts $DE$ at $T$. It is enough to show that $H$ halves $TE$.

Check this link

Since each $D$ and $E$ maps to it self, the circle $MDE$ maps to line $DE$ and so inversion swaps $K$ and $H$, where $K$ is intersection of line $MC$ and circle $MDE$. Also, this inversion swaps $F$ and $T$.

Since $\triangle CEK \sim \triangle CME \Longrightarrow \displaystyle{KE\over ME} = {CK\over CE} \;\;(1)$.

Similary $\triangle CFK \sim \triangle CMF \Longrightarrow \displaystyle{KF\over MF} = {CK\over CF} \;\;(2)$.

Now remeber that if $X',Y'$ are images of $X,Y$ we have $$X'Y' = XY \cdot {r^2\over XM\cdot YM}$$

So $$ HE = KE \cdot {r^2\over KM\cdot EM} \stackrel{(1)}{=}{CK\over CE} \cdot {r^2\over KM}$$ and So $$ HT = KF \cdot {r^2\over KM\cdot FM} \stackrel{(2)}{=}{CK\over CF} \cdot {r^2\over KM}$$

Since $CE = CF$ we have $HE= HT$. q.e.d.

Answer 3

Projective solution:

Let $t$ be a tangent in $M$. Then $t||DE$. Let $CM$ cuts segment $FE$ in $O$. Now we have:

\begin{eqnarray*} D(E,T;H, \infty) &=& D(ME,MT;MH, M\infty)\\ &=& D(ME,MF;MK, MM)\\ &=& D(E,F;K,M) \\ &=& D(EE,EF;EK,EM) \\ &=& D(EC,EO;EK,EM) \\ &=& D(C,O;K,M) \\ &=&-1 \end{eqnarray*} The last one is true since $O$ is on polar of $C$, so they are harmonical conjugate.

So $H$ halves $ET$ since $D(E,T;H, \infty)=-1$. In the same manner we prove that $G$ halves $DT$ and we are done.