## Saturday, April 18, 2009

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

Proposed Problem

See also: Complete Problem 282, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

Tr BAF gives,sin(α)= x/AB and from Tr.BAD, cos(α)= AB/AD. Putting these values in (1), we've:
Ajit: ajitathle@gmail.com

2. 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/2

P.S. angle @ mean alpha

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

Also < 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

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

Further ABJE is also cyclic

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

5. Problem 282
Consider 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

6. ABDE lies on a circle. Find the midpoint O of the line AD.

O is the center of the cirlce throught ABDE.
From O draw a line perpendicular onto the line DE. Call the line X.
Than we have :

DX =a/2

Join OB. Then the triangles OXD and OBF are congruent and we have

BF=XD
x=a/2