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Geometry Difficulty 5.5 AIME, harder Prove it Ukraine

Let ABCDABCD be a cyclic quadruple. Let us denote the midpoints of ABAB, BCBC, CDCD and DADA by MM, LL, NN and KK respectively. It is known, that ∠BMN=∠MNC\angle BMN = \angle MNC. Prove that:

a) ∠DKL=∠CLK\angle DKL = \angle CLK;

b) ABCDABCD has a pair of parallel sides.

Solution

a) Using the properties of inscribed angles we get KM⊥BDKM \perp BD, KN⊥ACKN \perp AC, ∠ABD=∠ACD⇒∠AMK=∠ABD=∠ACD=∠KND\angle ABD = \angle ACD \Rightarrow \angle AMK = \angle ABD = \angle ACD = \angle KND. Thus ∠KMN=π−∠AMK−∠BMN=π−∠KND−∠MNC=∠KNM\angle KMN = \pi - \angle AMK - \angle BMN = \pi - \angle KND - \angle MNC = \angle KNM, hence □KMN\square KMN is isosceles and KM=KNKM = KN, which implies that KMLNKMLN is rhombus and ∠NKL=∠NLK\angle NKL = \angle NLK (fig. 8). Since by analogy we have that ∠AMK=∠KND\angle AMK = \angle KND, ∠DKN=∠NLC\angle DKN = \angle NLC, then ∠KLD=∠DKN+∠NKL=∠NLC+∠LNK=∠CLK\angle KLD = \angle DKN + \angle NKL = \angle NLC + \angle LNK = \angle CLK, which proves part a).

b) Since KM=KNKM = KN, then DB=2KM=2KN=ACDB = 2KM = 2KN = AC and ∠ABC+∠DAB=π\angle ABC + \angle DAB = \pi. We have DA⊥BCDA \perp BC, or ∠ABC=∠DAB\angle ABC = \angle DAB. But since ∠DAC=∠DBC\angle DAC = \angle DBC, then ∠CAB=∠DAB−∠DAC=∠ABC−∠DBC=∠ABD=∠ACD\angle CAB = \angle DAB - \angle DAC = \angle ABC - \angle DBC = \angle ABD = \angle ACD, and AB⊥BCAB \perp BC.

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