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Geometry ProblemClick the figure below to see the complete problem 499 about Triangle, Two Squares, Perpendicular.
Let Q is projection of E on AC ; H is the projection of B on AC; P is the projection of F on ACLet O is the point of intersection of AF to BH and O’ is the point of intersection of CE to BH.We will prove that HO=HO’.Tri. AHO similar to tri. APF and we have HO/FP=AH/APReplace AH=c.cos(A) ; AP= AC+CP= b+a.cos(90-C)=b+a.sin(C) ; FP=a .cos(C)We get HO=a.c.cos(A).cos(C)/(b+a.sin(C))Similarly Tri. CHO’ similar to CQE and we have HO’/EQ=CH/CQReplace CH=a. cos(C) ; CQ=b +c.sin(A) ; EQ=c.cos(A) We get HO’= a.c.cos(A).cos(C)/(b+c.sin(A))Note that a,sin(C )=c.sin(A)=BH Replace this value in above formulas for HO and HO’ we will get HO=HO’. Point O coincide with O' and BO will perpendicular to AC at HPeter Tran
Any other, more direct proof?
Problem 499 (to Anonymous): Hints:Construct the external square ACPQ and prove that lines AQ and BO are parallel.More hints: prove that angles AQB and ABO are congruent, see problem 496, cyclic quadrilateral.
Construct square ACPQ BQ and BP cross EC and AF at N and M respectively. Now, triangles EAC and BAQ are congruent (SAS), EC is perpendicular to QB. Also, AN bisects the right angle (see problem 496).In the same way triangles FCA and BCP are congruent; BP is perpendicular to AF and CM bisectes the right angle.Let angle OBN= aQuad BNOM is cyclic (opp angles supplimentary)So angle NMO= a (angles on the cirumference subtended by same chord)Quad ANMC is cyclic (angles ANC, AMC are both 135)Angles NMA=NCA= a (angles on the cirumference subtended by same chord)Angle NCA =NQA= a (from congruency)Hence angle OBQ= angle BQA = a BO is parallel to AQBO is perpendicular to AC