Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Details: Click on the figure below.
Problem 1305Let AF, DG are perpendicular to BK, AB respectively.Then <AGD=<AHD=<AFD so the points A,G,D,F and H are concyclic .Hence <GFB=<GFD=<GAD=<BAE=<CAE=<CBE, so GF//BE. Similar HF//EC.If the BC intersects GE and HE at K,L respectively .Let area triangle ABC is (ABC) then (BGK)=(FKE) and ((HLC)=(FLE).(ABC)=(AGKLH)+(BGK)+(HLC)=(AGKLH)+(FKE)+(FLE)=(AGEH)=2.(AEH).APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE
https://goo.gl/photos/2Wk2e5vJSCUmWPte8Draw DF ⊥ ABTriangle AHE congruent to AFE ( case SAS) ( see sketch)FH⊥AE=> Area(AFEH)=AE.MH…= 2.Area(AHE)..(1)Applying Ptolemy’s theorem in quad. ABEC give us AE= (b+c).BE/BCNote that BE/BC= AD/(2.AH) and AD.HM=AH.DH ( relation in right triangle AHD )Replace these values to (1) we will have Area(AFEH)=1/2. DH(b+c)Now Area(ABC)=Area(ADB)+Area(ADC)= ½.DH(b+c)Comparing these 2 areas we will have Area(AHE)= ½.Area(ABC)
Let DH = h, altitude of ∆AHD from H be p, AE = q and BE = CE = d.From Ptolemy qa = d(b+c) ….(1)S(ADC) = ½ hb = b/(b+c).S(ABC)…(2)S(AHE) = ½ pq….(3)p/h = ½ a/d …(4) from similar ∆s.(3)/(2) => S(AHE) /S(ABC). (b+c)/b = pq/(hb) = ½ qa/(bd) from (4)Now substituting for qa/d from (1),S(AHE)/S(ABC) . (b+c)/b = ½ (b+c)/bHence S(AHE) = ½ S(ABC)Sumith PeirisMoratuwaSri Lanka
Join BE and form the triangle ABEm(AEB) = m(ACB) (angles in same segment) --------(1)m(BAE) = m(EAC) = x (say) ----------(2) (given)So triangles AEB and ACD are similar=> AB/AE = AD/AC=> AB.AC = AE.AD ----------(3)Area of triangle AEH = 0.5*AE*AH*sin(x)=> AEH = 0.5*AE*AD*cos(x)*sin(x)=> AEH = 0.5*0.5*AE*AD*sin(2x)=> AEH = 0.5*0.5*AB*AC*sin(2x) (from (3))=> AEH = 0.5*(Area of ABC)
through AE, reflect HE,get H2E, H2 on AB;connect DH2;area AHEH2 = 2x area AHEconnect HH2, AE perpendicular to HH2;Through AO make diameter AK of circle O;connect KH2, DK,HK,also connect KB,KC,KE,KB//DH2; KC//DH; KE//HH2;area BDH2 =area DKH2, (KB//DH2); area DHC= area DHK, (KC//DH);area KHH2 = area EHH2, (KE//HH2)so area ABC =area AHKH2 =area AHEH2 =2x area AHE