A Cyclic Quadrilateral
Ken Holing from Norway sent the following exercise:
Let I be the incenter of a triangle ABC and let D be the point of contact of the incircle and side AB. If ID is extended outside AB to H such that DH is equal to the semiperimeter of the triangle ABC, show that the quadrilateral AHBI is cyclic if and only if the angle C is right. Let a, b, and c be the lengths of the sides BC, AC, and AB respectively.
We will use a LEMMA:
Given s is the semiperimeter for a triangle with sides of length a, b, and c,
s(s - c) = (s - a)(s - b)
if and only if the triangle is a right triangle and c is the hypotenuse.
See the algebra exercise for s(s - c) = (s - a)(s - b).
Construct a triangle ABC and its incenter with center I and point of tangency D on AB.
Likeswise denote E on BC and F on AC, the points of tangency of the incircle on those sides. Now let
But, we also have
Next, construct the circle through A, I, and B. Extend ID to its intersection K with the circle.
We will show that if angle C is a right angle
then the length of DK = s.
If angle C is a right angle, the IF = ID = IE = z = s - c.
Angles BAI and BKI are equal since they subtend the same chord BI. Angles ADI and KDB are vertical angles. Therefore triangle ADI and KDB are similar. We have:
By the Lemma, we know that since ABC is a right triangle, KD = s
On the other hand, suppose we have constructed the extension HD = s on side AB and that AHBI is cyclic.
Angles BHI and BAI are equal, subtending the
same chord BI. Angles ADI and HDB are vertical angles. Therefore
triangles ADI and HDB are similar. We have,
From the Lemma, we know that when ID = s - c we have a right triangle with a right angle at C.
CHALLENGE: Find alternative proofs of this theorem that the quadrilateral is cyclic if and only if the angle at C is a right angle.
Note that when angle C is a right angle, the two diagonals of the quadrilateral are