The problem is equivalent to $\angle EOH=45^\circ$, where $O$ is the center of $ABCD$. Try it yourself, it's easy.

yes of corse it's equivalent to $\widehat{EOH}=45^{\circ }$ But how can you prove that ?

erdos wrote: yes of corse it's equivalent to $\widehat{EOH}=45^{\circ }$ But how can you prove that ? Let X,Y be the midpoints of AB and BC respectively. Then our goal is to show EH=EX+HY. WLOG, assume AB=BC=CD=DA=2. Let EX=x. Then by similar triangle, $\frac{1}{x}=\frac{CH}{HB}$. Thus $BH=\frac{2x}{1+x},HY=\frac{1-x}{1+x}$. I think it's simple enough now

Let $L$ be the midpoint of $AB$, and $O$ the midpoint of $LF$. Consider $L(0,0)$ the origin, $LO=1$, and $E(0,t)$. The equation of the circle centered at $O$ and radius $1$ is $(x-1)^2+y^2=1$ Then using the given $ {ED}||{HF}$ we can easily find $H\left(\frac{2t}{t+1},1\right)$. The line $l$ through $E$ and $F$ has the equation $y=\frac{1-t^2}{2t}x+t$. Consider the point $K\left(\frac{2t^2}{t^2+1},\frac{2t}{t^2+1}\right)$. Then an easy computation shows that $K$ belongs to $l$ and to the circle centered at $O$ and radius 1. It remains to see that $l_1$ (the line through $K$ and $O$) has the angular coefficient $\frac{2t}{t^2-1}$ hence ${l_1}\perp{l}$ and we conclude that $l$ is tangent to the circle inscribed in $ABCD$ at $K$.

Here's what I would call a more explanatory solution of the original problem. Problem. In a square ABCD, let F be the midpoint of the side CD, and let E be a point on the line AB. The parallel to the line DE through the point F meets the line BC at a point H. Prove that the line EH is tangent to the incircle of the square ABCD. Solution. Consider the two tangents to the incircle of the square ABCD through the point E. One of these two tangents is clearly the line AB. Let the other tangent intersect the line BC at a point H'. We will show that DE || FH'. This will solve the problem. Why? Well, once it is known that DE || FH', it follows that the point H' lies on the parallel to the line DE through the point F; in other words, the point H' is the point of intersection of the line BC with the parallel to the line DE through the point F. But we have defined the point H as the point of intersection of the line BC with the parallel to the line DE through the point F. Thus, H' = H. Since we know that the line EH' is tangent to the incircle of the square ABCD, we can thus conclude that the line EH is tangent to the incircle of the square ABCD. And the problem is solved. So it remains to show that DE || FH'. Before we show this, we cite the Brianchon theorem in its "real" form (i. e., in its most general form): Brianchon theorem. Let k be a circle and $t_1$, $t_2$, $t_3$, $t_4$, $t_5$, $t_6$ six tangents to this circle. Let $P_1=t_1\cap t_2$, $P_2=t_2\cap t_3$, $P_3=t_3\cap t_4$, $P_4=t_4\cap t_5$, $P_5=t_5\cap t_6$, $P_6=t_6\cap t_1$ be the pairwise points of intersection of "adjacent" pairs of these tangents; hereby, if two adjacent tangents (e. g., the tangents $t_3$ and $t_4$) coincide, then their point of intersection is defined as the point where they are tangent to the circle k. Then, the lines $P_1P_4$, $P_2P_5$ and $P_3P_6$ are concurrent. Before we apply this Brianchon theorem to our square, we make some preliminary observations: First of all, the point F, being the midpoint of the side CD of the square ABCD, must be the point of tangency of the incircle of this square with the side CD (this simply follows from the symmetries of the square). Also, since ABCD is a square, the lines BC and DA are parallel; thus, they have a common infinite point X. Similarly, the lines AB and CD have a common infinite point Y. Now consider the incircle of the square ABCD and the six lines AB, EH', BC, DA, CD, CD which are all tangents to this incircle. We want to apply the Brianchon theorem to these six tangents. We have $AB\cap EH^{\prime}=E$, $EH^{\prime}\cap BC=H^{\prime}$, $BC\cap DA=X$, $DA\cap CD=D$, $CD\cap CD=F$ (here we use the convention from the Brianchon theorem that if two adjacent tangents coincide, then their point of intersection is defined as the point where they are tangent to the circle k - in our case, the two coinciding tangents are CD and CD, and thus, their point of intersection is the point of tangency of the incircle of the square ABCD with the line CD, i. e. the point F) and $CD\cap AB=Y$. Thus, after the Brianchon theorem, the lines ED, H'F and XY are concurrent. In other words, the point of intersection of the lines ED and H'F lies on the line XY. But since the points X and Y are infinite points, the line XY is the line at infinity; consequently, the point of intersection of the lines ED and H'F lies on the line at infinity, i. e. we have ED || H'F. In other words, DE || FH'. Proof complete. darij

Dear Mathlinkers, you can also see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=373081 Sincerely Jean-Louis

Dear Mathlinkers, look at http://jl.ayme.pagesperso-orange.fr/ vol. 7, Miniatures géométriques, probleme 12 p. 24. Sincerely Jean-Louis