Problem: Let X be the point on side BC such that BX=CD. Show that the excircle ABC opposite of vertex A touches segment BC at X.
Solution
Solution: Let the excircle touch lines BC, AC and AB at X′, Y and Z, respectively. Using the equal tangent property repeatedly, we have BX′−X′C=BZ−CY=(EY−CY)−(FZ−BZ)=CE−BF=CD−BD. It follows that BX′=CD, and thus X′=X. So the excircle touches BC at X.
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Source: MathNet,
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