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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1257.
Join FD, quad. ACDF is cyclic. AngleGFD = C, Also quad. BFEC is cyclic.Hence Angle AFE=Angle JFG= C. Angle FGD = Angle FGJ = 90By ASA Tr.FGD is congruent to Tr. FGJ, hence JG=GDQuad GHDB is cyclic Angle GDH =90-AAnd Angle GDB = 90-B, Angle HDB=90-A +90-B=C. We get Angle AGL=CSince Angle AGL = Angle AFE, JP is parallel to GL, in Tr. JPD, G is midpoint of JD and GL is parallel to JP, hence JP=2×GLJK + PK =2×GLJK =2×10-9=11
https://goo.gl/photos/6syuJ3KQA1nGxwFk7Let N is the orthocenter of triangle ABCWe have cyclic quadrilaterals AENF, ECDN, ACDF, BGHD and AEDB ( see sketch)Let u= ∠ (EBC)= ∠ (DAC)= ∠ (HGD)= ∠ (FEN)Let v= ∠ (BAD)= ∠ (BCF) and z=∠ (ABE)= ∠ (ACF)We have ∠ (AFE)= ∠ (JFG)= 90-u => ∠ (FJG)= u => JE//GLIn right triangle ADC we have u+v+z= 90In right triangle ADG we have ∠ (FDG)+v+z= 90 => ∠ (FDG)= uTriangle FJD is isosceles => DG/DJ=DL/DP= ½So JP= 2. GL= 20 => KJ= JP- KP= 11
Problem 1257Let O is orthocenter the triangle ABC, the quadrilateral AFOE,EODC and AFDC are cyclic so <FEO=<FAO=<FAD=<FCD=<OCD=<OED and triangles KHE ,DHE are equals(<EHK=<EHD=90)soKH=DH.If EF intersects the circle passing through the points A,E,Dand B(<AEB=90=<ADB) in J’ then arc J’B=arcBD(J’B=BD , <KEH=<DEH=J’EB=<BED).But the diameter of the circle (J’,A,E,D,B) is AB so J’D is perpendicular bisector in AB.Then the points J,J’ coincide so JG=GD.Therefore GL//JP and PJ=2GL=2.10=20.But JK=JP-KP=20-9=11.APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
We have a number of cyclic quads and accordingly < B = < GHK = < AEF = < EKDHence GL // JPBut also < B = < DEC = < FEA hence Tr.s KHE and DHE are isoceles and KH = HD.Hence from the midpoint theorem HL = 9/2 and so GH = 11/2 and therefore JK = 11Sumith PeirisMoratuwaSri Lanka
CorrectionTr.s KHE and DHE are congruent not isoceles
A shorter cut would have been to realize that if O is the orthocentre of Tr. ABC then O is the incentre of Tr. DEF yielding both tr.s JFD and DEK isoceles. Either result can then be used along with the mid point theorem to get the value of JKSumith PeirisMoratuwaSri Lanka