beauty of geometry
Geometry
Level
pending
) In a triangle ABC satisfying AB+BC = 3AC the incircle has centre I and touches the sides AB and BC at D and E, respectively. Let K and L be the symmetric points of D and E with respect to I.Quadrilateral ACKL is ?
complete
kite
square
cyclic
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1 solution
Let F be the point of tangency of the incircle with AC and let M and N be the respective points of tangency of AB and BC with the corresponding excircles. If I is the incenter and Ia and P respectively the center and the tangency point with ray AC of the excircle corresponding to A, we have Al/IL =AI/IF =AIa/IaP =AIa/IaN , which implies that △AIL ∼ △AIaN. Thus L lies on AN, and analogously K lies on CM. Denote x = AF and y = CF. Since BD = BE, AD = BM = x, and CE = BN = y the condition AB+BC = 3AC gives us DM = y and EN = x. Now the triangles CLN and MKA are congruent since their altitudes KD and LE satisfy DK = EL, DM =CE, and AD = EN. Thus ∠AKM = ∠CLN, implying that ACKL is cyclic.