Online Geometry theorems, problems, solutions, and related topics.
Proposed ProblemSee also: Complete Problem 282, Collection of Geometry ProblemsLevel: High School, SAT Prep, College geometry
In Tr DAE, a=AD*sin(2α)=2AD*sin(α)*cos(α)---(1)Tr BAF gives,sin(α)= x/AB and from Tr.BAD, cos(α)= AB/AD. Putting these values in (1), we've:a=2*AD*x/AB*AB/AD or a =2x or x=a/2Ajit: firstname.lastname@example.org
draw BO, O midpoint of AD, => BO = OA = OD as radius of circle through A,B,C (B=90)angle FBD = angle BAD =@,(perpendicular sides)=>angle BOD = 2@ ( tr. BOD has BO=DO )=> tr BFO and ADE are similar => x/BO = a/AD, but AD=2r, BO=r=> x/r = a/2r, => x = a/2P.S. angle @ mean alpha
Let ths diagonals of the cyclic quad ABDE meet at G and let H be ths midpoint of GEAlso < GBD = < DAE so BF bisects < DBG and F is the midpoint of GD. So from midpoint theorem FH = a/2. Also FH is // to DE hence < FHB = < GDE = < BAD = < DBF = < FBG showing that Tr. BFH is isoceles Hence x = FH = a/2Sumith PeirisMoratuwaSri Lanka
Observe that AD is the diameter of cyclic quad ABDE and AB is the diameter of cyclic quad ABFHFurther ABJE is also cyclic If BF meets AH at J, then Tr. s AGE, BGD, BFH, HFJ, ABJ are all isoceles
Problem 282Consider that BE intersect Ad in N and K is symmetric of B with respect to AD.Then ABDE is cyclic <DBE=2α so <EBF=α.Therefore BDKN is rhombus (ΒΝ=//ΔΚ ) and <AKD=90 .Then ABDK is cyclic.But <AEB=<ADB=<BND=<ANE so AN=AE.Is triangleADK=triangleAKM (DK intersect AC in M).Now DK=KM and BK=//KM so NMKB is parallelogram and 2.BF=BK=NM.But BK is perpendicular in AD,NM is perpendicular in AD .Now AD=AM(<ADM=<ANE=<AEN=<AMD ).So DE=MN=BK=2BF.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE