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.Click the figure below to view more details of problem 1255.
Problem 1255Solution 1In accordance with problem 848 in triangle BCD <BDC=26.5 so <BCD=90-26.5=63.5.But accordance with problem 849 in triangle CDE <DEC=18.5 so <DCE=90-18.5=71.5.Now <ACB+<BCD+<DCE=45+63.5+71.5=180.Therefore A,C, and E are collinear.Solution 2Let the points K,L on the side DE such that DK=KL=LE. If <DKC=α, <DLC=β and <DEC=θthen in accordance with problem 363 in triangle DCE apply α+β+θ=90 or β+θ=45.Is triangle BCD similar with triangle DCL(BC=2BD, DC=2DL) so <BCD=<DCL=90-β and<DCE=90-θ. But <ACB+<BCD<DCE=45+(90-β)+(90-θ)=45+180-(β+θ)=45+180-45=180.Therefore A,C, and E are collinearAPOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
Geometry solution:https://goo.gl/photos/sdw3M9H3Ktv2F8PVALet x=AB=BC and BD=2xLet AC cut circumcircle of triangle BDC at FNote that ADF is 45-45-90 triangle soDF=AF=3x/√(2)CF=AF-AC=3x/√(2) – x√2= x/√(2)And CD/DE=CF/DF=1/3Triangle DFC similar to EDC ( case SAS)So angle (DCF)= angle (DCE)=> A, C and E are collinear.Trigonometry solution:Let u= ∠ (BCD) and v= ∠ (DCE)We have ∠ (BCE)=u+vSo tan(∠ (BCE))= (tan(u)+tan(v))/(1- tan(u).tan(v))= (2+3)/(1-2.3)= -1So ∠ (BCE=135 degrees and ∠ (ACE)= 45+135= 180 degreesSo A,C and E are collinear
Geometry SolutionDrop a perpendicular EF from E to AF.Let AB = BC = p and CD = q so that BD = 2p and DE = 3qNow triangles CBD and DFE are similar hence DE = 3p and FE = 6p since DE = 3DC. So AF = 6p = FE hence < FAE = 45. But < FAC = 45It thus follows that A,C,E are collinearSumith PeirisMoratuwaSri Lanka
Trigonometry SolutionLet < BCD = m and let < DCE = nSo tan(m+n) = (2+3)/(1-2X3) = -1Hence m+n = 135 degrees and the result followsSumith PeirisMoratuwaSri Lanka
Let AB = BC = 1 => AC = Sqrt(2)BD = 2 => DC = Sqrt(5)Join CE and form the Right triangle CDECD = Sqrt(5)DE = 3*Sqrt(5)CE = 5*Sqrt(2)Drop a _|_ from D to EC that meets at F, DF = 3/Sqrt(2) Area of ADC = 0.5*AD*BC = 0.5*3*1 --------(1)Area of DCE = 0.5*DC*DE = 0.5*Sqrt(5)*3*Sqrt(5) = 0.5*5*3 -------------(2)Adding (1) & (2) = 9 Sq.units ---------(3)Let us assume A,C & E are collinearJoin AE and form the triangle ADE. Area of ADE = 0.5*AE*DF = 0.5*6*Sqrt(2)*3/Sqrt(2) = 9 Sq.units ------------(4)Since (3) = (4), our assumption is true and A,C & E are collinear