[alogo] Bisector rectangle

Consider triangle ABC and its circumcircle c. For a point A* moving on c, construct triangle A*BC and the bisectors AF and A*G of the two triangles. Consider then the circles inscribed and escribed in angles A and A*. Their centers form a rectangle EFGH.

[0_0] [0_1] [0_2]
[1_0] [1_1] [1_2]

The proof is direct consequence of the property proved in Bisector.html . I is the intersection point of the two bisectors and coincides with the middle of the arc BC. The six points B, F, G, C, H and E are equidistant from I. Notice that the angle(FIG) is equal to angle B of the triangle. Hence, varying C on (c) the rectangle EFGH changes its dimensions but remains similar to itself.

Considering the other pairs of triangles, like (BCA, BCA*) in the cyclic quadrangle BCA*A one can construct an interesting figure containing a cross. Look at the file BisectorCross.html for a further discussion on this.

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