Proving the lemma when the intersection is inside. In our book we had a lemma that says if $G$ and $H$ are tangent points and $F$ and $E$ are mid points then $GH$,$FE$ and $CD$ intersect in the same point and the angle $BIC=90$.

This is the lemma when the intersection is outside.To prove this lemma we can connect $D$ to $H$ and get a right angle then we have to prove $DHIB$ is cyclic which can easily proved by showing that angles $DBH=HID$.When the intersection is inside proving $BIC=90$ is a bit different but possible using cyclic quadrilaterals but then I don't know how to prove these three lines intersect in the same point.
 A: *

*Let the line $CD$ intersect $EF$ at point $I_1$. $\, EF$ is a midsegment so   $EF || AC$ and $EF = \frac{1}{2} \, CA$. Thus $EI_1 || CA$. Since $CD$ is an angle bisector and  $EI_1 || CA$
$$ \angle \, ECI_1 = \angle \, ACI_1  = \angle \, EI_1C$$ which means that triangle $CEI_1$ is isosceles so $$EI_1 = CE = BE$$ which by the way means that triangle $CBI_1$ is right-angled. 


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*Let line $GH$ intersect $EF$ at point $I_2$. Since $EI_2 || CA$ triangles $AGH$ and $FI_1H$ are similar and since $AG = AH$, also $FI_1 = FH$. Our goal is, given that $BC = a, \, CA = b, \, AB = c$ to calculate the length of $$EI_2 = EF + FI_1 = \frac{1}{2} \, CA + FI_1 = \frac{b}{2} + FI_1 = \frac{b}{2} + FH$$ Thus, we need to calculate $FH$. But $$FH = FA - AH = \frac{1}{2} \, AB - AH = \frac{c}{2} - AH$$ Moreover, 
$$AH = \frac{b+c - a}{2}$$ so 
$$FH =\frac{c}{2} - AH =  \frac{c}{2} - \frac{b+c- a}{2} = \frac{a-b}{2}$$ Therefore, $$EI_2 = EF + FI_1 = EF + FH = \frac{b}{2} + \frac{a-b}{2} = \frac{a}{2} = \frac{1}{2} \, BC = CE$$ However, we proved already that $EI_1 = CE$. Therefore $EI_1 = EI_2$ which is possible if and only if $I_1 \equiv I_2 $. Thus the tree lines $GH, \, CD$ and $EF$ intersect at a common point we call $I \equiv I_1 \equiv I_2$. 


We already proved that $\angle \, BIC \equiv \angle \, BI_1C = 90$
