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1/XT=1/XY + 1/ XZ wanted, start with 2 intersecting circles, 2 more circles

Source: OLCOMA Costa Rica National Olympiad, Final Round, 2014 4.1

September 24, 2021
geometrycircles

Problem Statement

Let A A and B B be the intersections of two circumferences Γ1\Gamma_1, and Γ2\Gamma_2. Let CC and DD points in Γ1\Gamma_1 and Γ2\Gamma_2 respectively such that AC=ADAC = AD. Let EE and FF be points in Γ1\Gamma_1 and Γ2\Gamma_2, such that ∠ABE=∠ABF=90o\angle ABE = \angle ABF = 90^o. Let K1K_1 and K2K_2 be circumferences with centers EE and FF and radii ECEC and FDFD respectively. Let TT be a point in the line ABAB, but outside the segment, with T≠AT\ne A and T≠A′T \ne A', where A′A' is the point symmetric to AA with respect to B B. Let XX be the point of tangency of a tangent to K1K_1 passing through TT, such that there arc two points of intersection of the line TXTX to K2K_2. Let YY and ZZ be such points. Prove that 1XT=1XY+1XZ.\frac{1}{XT}=\frac{1}{XY} + \frac{1}{XZ}.
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