Proving that the circumcircle of triangle is tangent to the line

Problem 5:

Let ABCD be a convex quadrilateral with $\angle B=\angle D=90^0$. Point H is the foot of the perpendicular from A to BD. The points S and T are chosen on the sides AB and AD, respectively, in such a way that H lies inside triangle SCT and

$\angle SHC=\angle BSC=90^0$, $\angle THC=\angle DTC=90^0$

Prove that the circumcircle of triangle SHT is tangent to the line BD.

Solution.

Let the line passing through C and perpendicular to the line SC intersect the line AB at Q (see Figure 1). Then

$\angle SQC=90^0-\angle BSC=180^0-\angle SHC$,

which implies that the points C, H, S, and Q lie on a common circle. Moreover, since SQ is a diameter of this circle, we infer that the circumcentre K of triangle SHC lies on the line AB. Similarly, we prove that the circumcentre L of triangle CHT lies on the line AD.

In order to prove that the circumcircle of triangle SHT is tangent to BD, it suffices to show that the perpendicular bisectors of HS and HT intersect on the line AH. However, these two perpendicular bisectors coincide with the angle bisectors of angles AKH and ALH. Therefore, in order to complete the solution, it is enough (by the bisector theorem) to show that

$\frac{AK}{KH}=\frac{AL}{LH}$ ………………(1)

We present two proofs of this equality.

First proof. Let the lines KL and HC intersect at M (see Figure 2). Since KH = KC and LH = LC, the points H and C are symmetric to each other with respect to the line KL. Therefore M is the midpoint of HC. Denote by O the circumcentre of quadrilateral ABCD. Then O is the midpoint of AC. Therefore we have OM || AH and hence $OM\perp BD$. This together with the equality OB = OD implies that OM is the perpendicular bisector of BD and therefore BM = DM.

Since CM K KL, the points B, C, M, and K lie on a common circle with diameter KC.Similarly, the points L, C, M, and D lie on a circle with diameter LC. Thus, using the sine law, we obtain

which finishes the proof of (1).

Second proof. If the points A, H, and C are collinear, then AK = AL and KH = LH, so the equality (1) follows. Assume therefore that the points A, H, and C do not lie in a line and consider the circle ω passing through them (see Figure 3). Since the quadrilateral ABCD is cyclic,

$\angle BAC=\angle BDC=90^0-\angle ADH=\angle HAD$.

Let $N\neq A$ be the intersection point of the circle ω and the angle bisector of $\angle CAH$. Then AN is also the angle bisector of $\angle BAD$. Since H and C are symmetric to each other with respect to the line KL and HN = NC, it follows that both N and the centre of ω lie on the line KL. This means that the circle ω is an Apollonius circle of the points K and L. This immediately yields (1).

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