# A Property of Circumscribed QuadrilateralsWhat is it? A Mathematical Droodle

Explanation  Assume the circle inscribed into a quadrilateral ABCD touches its sides at points E, F, G, H, as on the diagram. Then the lines EG and FH pass through the point M, the intersection of diagonals.

To prove the fact we shall assume that N is the point of intersection of AC and EG. The most basic property used in the proof is the fact that the two tangents at the endpoints of a chord form equal angles with that chord. Thus, for example,

(1)

∠AEN = ∠DGN.

Further, supplementary angles, i. e. angles that add up to 180°, have the same sine. For example,

(2)

sin(∠AEN) = sin(∠DGN) = sin(∠CGN).

An additional fact of importance is that the area of a triangle could be computed as half the product of two of its sides times the sine of the angle formed by the two sides. For ΔAEN we have,

(3)

2·Area( ΔAEN) = AN·EN·sin(∠ANE) = AE·EN·sin(∠AEN).

For ΔCGN we have,

(4)

2·Area( ΔCGN) = CN·GN·sin(∠CNG) = CG·GN·sin(∠CGN).

Let's combine (1)-(4):

Area( ΔAEN)/Area( ΔCGN) = AN·EN/CN·GN = AE·EN/CG·GN,

from where

AN·EN/CN·GN = AE·EN/CG·GN,

or

(5)

AN/CN = AE/CG,

which means that the line EG divides the diagonal AC in the ratio AE/CG.

By a similar argument, now assuming that AC and FH meet at some point, say N', and considering triangles AHN' and CFN', we show that FH divides AC in the ratio AH/CF. However, AE = AH and CG = CF as the tangents from the same point to a circle. From here, the lines EG and FH divide AC in the same ratio. In other words, N and N' are one and the same point. The diagonal AC passes through the point of intersection of EG and FH. Invoking similarity once again we may claim that the same is true of the diagonal BD.

The four lines are concurrent at N, but then N = M, the point of intersection of the diagonals.

### Remark

The theorem is also seen as a consequence of Brianchon's and Ceva's theorems, so that it is actually true for more general conic sections, not just circles. (The hexagon ABFCDH and AEBCGD are circumscribed around a circle. Therefore Brianchon's theorem applies. The point of intersection of AC and BD being a fixture in both cases both EG and FH pass through that point.)

Also, the theorem can be shown based on the nature of the relationship between poles and polars and by a duality argument. 