Inscribed Circles in a Quadrilateral

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I found this problem online, where it was asked to prove EF = GH.

enter image description here

I was able to prove that, but got intrigued by how the four smaller inscribed circles could be constructed in the first place.

That is, given the fact that AD + BC = AB + CD (i.e. an inscribed circle can be constructed for ABCD), how can we contruct EF and GH, such that for each smaller quadrilateral, an incribed circle can be constructed?

Also, given a fixed quadrilateral with AD + BC = AB + CD, are the positions of EF and GH unique?

Any hints would be appreciated.

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There are 2 best solutions below

8
On

enter image description here

Hint:

Use this fact:

ُThe center of four circle I, J, K and L are on a circle (M) concentric with inscribed circle N at O and Lines AO, BO, CO and DO which connect vertices of quadrilateral to the center (O) of inscribed circle.

Now follow this rule:

1- Draw AO. BO, CO and DO.

2-draw AC and BD, they intersect on P.

3-Connect OP.

4- find the midpoint of OP and mark it as Q. Q is the intersection of EF and GH (in your figure P).

5- Draw a line parallel with BD, it intersect BO and DO on J and L respectively.these are the center of two opposite circles.They also define the measure of the radius (M).

6- Draw circle (M), it intersect AO and CO on I and K respectively. these are the centers of other two opposite circles.

7- Draw four cicles.

8- draw common tangents of adjacent circles, they will intersect on Q( in your figure P) and you get EF and GH.

The reason for taking Q as the midpoint of OP is that if equilateral is regular, points P and O and Q will be coincident. If it is not regular , they will have a distance like OP and the intersection of common tangents locates on its midpoint Q.

6
On

We're going to use

Graves-Chasles Theorem (G-C) $\quad$ Let $ABCD$ be a convex quadrilateral such that all its sides touch a conic $\alpha$. Then the following four properties are equivalent:

  1. There exists a circle inscribed into $ABCD$.
  2. The points $A$ and $C$ lie on a conic confocal with $\alpha$.
  3. The points $B$ and $D$ lie on a conic confocal with $\alpha$.
  4. The intersections of opposite sides lie on a conic confocal with $\alpha$.

Graves-Chasles

Let the foci $F_1,F_2$ be the intersections $AB\cdot CD$ and $AD\cdot BC$ respectively. We let conic $\alpha$ be the degenerate ellipse with foci $F_1,F_2$ and eccentricity $1$. By G-C, $A,C$ lie on an ellipse $\beta$ (blue) confocal with $\alpha$, and $D,B$ lie on a hyperbola $\gamma$(green) confocal with $\alpha$.

Let $P$ be the intersection of $\beta$ and $\gamma$, and let the lines $PF_1$ and $PF_2$ cut the quadrilateral in $E,F,G,H$.

By G-C, the four sub-quads are tangential (i.e. have an incircle). For example, for $DEPH$, properties $3,4 \implies 1$.

Much more detail, including a proof of G-C, can be found in Akopyan & Bobenko, Incircular nets and confocal conics. Notably, this construction can be recursed and generalized to create so-called incircular nets. G-C is Theorem 2.5 in the paper.