Proposed Problem

See also: Complete Problem 282, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Saturday, April 18, 2009

### Problem 282: Right Triangle, Cevian, Angles, Perpendicular, Congruence

Labels:
angle,
cevian,
congruence,
perpendicular,
right triangle

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In Tr DAE, a=AD*sin(2α)=2AD*sin(α)*cos(α)---(1)

ReplyDeleteTr 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/2

Ajit: ajitathle@gmail.com

draw BO, O midpoint of AD, => BO = OA = OD as radius of circle through A,B,C (B=90)

ReplyDeleteangle 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/2

P.S. angle @ mean alpha

Let ths diagonals of the cyclic quad ABDE meet at G and let H be ths midpoint of GE

ReplyDeleteAlso < 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/2

Sumith Peiris

Moratuwa

Sri Lanka

Observe that AD is the diameter of cyclic quad ABDE and AB is the diameter of cyclic quad ABFH

ReplyDeleteFurther ABJE is also cyclic

If BF meets AH at J, then Tr. s AGE, BGD, BFH, HFJ, ABJ are all isoceles

Problem 282

ReplyDeleteConsider 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