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AE=BE, AF =CF, <BTE= <CTF=90^o, prove TA^2 =TB \cdot TC

Source: Dutch IMO TST2 2018 P3

August 5, 2019
geometryright angleequal segments

Problem Statement

Let ABCABC be an acute triangle, and let DD be the foot of the altitude through AA. On ADAD, there are distinct points EE and FF such that ∣AE∣=∣BE∣|AE| = |BE| and ∣AF∣=∣CF∣|AF| =|CF|. A pointT≠D T \ne D satis es ∠BTE=∠CTF=90o\angle BTE = \angle CTF = 90^o. Show that ∣TA∣2=∣TB∣⋅∣TC∣|TA|^2 =|TB| \cdot |TC|.
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