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Reciprocal geometric equation with excenter and circumcenter

Source: 2022 Turkey JBMO TST P7 + 2023 Dutch BxMO TST, Problem 4

March 15, 2022
geometrycircumcircle

Problem Statement

In a triangle △ABC\triangle ABC with ∠ABC<∠BCA\angle ABC < \angle BCA, we define KK as the excenter with respect to AA. The lines AKAK and BCBC intersect in a point DD. Let EE be the circumcenter of △BKC\triangle BKC. Prove that 1∣KA∣=1∣KD∣+1∣KE∣.\frac{1}{|KA|} = \frac{1}{|KD|} + \frac{1}{|KE|}.
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