See complete Problem 47
Angles, Triangle, Congruence. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Elearn Geometry Problem 47
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See complete Problem 47
Angles, Triangle, Congruence. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
good problem
ReplyDeletewe have to draw a perpendicular from from B to AC and E is the foot of the perpendicular .prove triangle BED is a rightangled isoscleles triangle
hence the result
Draw BE,CF perpendicular to AC & AB extended respectively. Let AD=DC=p. From Tr. ACF we've CF = AC/2 = p while angle FBC =30+15=45; hence BF=p and from Tr. ACF, AF=2p*V3/2 =V3*p. Thus, AB=V3p-p. Now angle ABE=60 and so BE=(V3p-p)/2 and AE=(V3p-p)V3/2 which gives DE = p-(V3p-p)V3/2 =(V3p-p)/2. In other words BE = DE or angle DBE=45 deg. Finally, angle ABE + angle DBE + x + 45 = 180 or 60 + 45 + 45 + x = 180 which yields x = 30 deg.
ReplyDeleteAjit: ajitathle@gmail.com
Prolong AB, then draw a perpendicular from C to AB, being F de foot of this perpendicular.
ReplyDeleteThen triangle DFC is euqilateral and BDF is isósceles of base BD.
Completing angles in the diferents triangles we'll have in BDF that 30 + 2(45+x)=180 <=>2x=60 <=> x=30
Greetings :D!
Drop a perpendicular CE to AB extended. Then Tr. BCE is right isocoles and BC^2 = 2BE^2 = CD.CA
ReplyDeleteHence BC is tangential to Tr. ABD and so x = 30
Sumith Peiris
Moratuwa
Sri Lanka
Or let E be the foot of the perpendicular from C to AB and let F be on AE extended such that BE = EF
ReplyDeleteThen AD = DC = DE = BE = CE = EF
So < BDF = 90 = < BCF. Hence BDCF is cyclic and < BFC = 45 = < ADB
Hence x = 30
Take E reflection of C about AB, tr ACE is equilateral, <CBE=<CDE=90, BDCE is cyclic, hence <CBD=<CED=30, done.
ReplyDeleteDraw equilateral triangle EBC ( E and A on the same side of BC)
ReplyDeleteE is the centre of the circumcircle of ABC. ang(EBA)=75 (135-60) ang(BAE)= 75;
From this: ACE is an isosceles triangle 45-45-90; ED=AD=DC ; ED=DC and EB=BC
BD bisects ang(EBC) -> x=30
Extend AB to a point E such that AE=EC => m(AEC)=120
ReplyDeleteand thus in triangle BEC, m(EBC)=45 -----(1)
Join ED and let F be the point of intersection of ED with BC
join AF and AFC is an isosceles triangle with m(FAC)=m(FAD)=m(FAB)=15 --------(2)
Let G be a point on AE such that AG=AD, hence m(AGD)=m(ADG)=75 --------(3)
Since m(GED)=60 and m(DGE)=105 (from (3))
=> m(GDE)=m(GDF)=15---------(4)
From (2) & (4) GFDA are concyclic => m(DGF)=15=> FD=FG and GDF is isosceles --------(5)
From (3) and (5) m(AGF)=m(BGF)=m(AGD)+m(DGF)=90 and hence BGF is an isosclese right triangle
=> BF=GF=FD and F is the circumcenter of BGD and BFD is isosceles
=>m(DBG)=m(BGD)=75
and m(DBF)=x=75-45=30
Mark E above AC such that DE=DC=AD and m(EDC)=60. Join AE and CE to form the right triangle AEC. since DE=DC and m(EDC)=60 => DEC is an equilateral triangle.
ReplyDelete=>m(ECD)=45 and since m(CDA)=90=> m(EBC)=45 => BE=EC=DE=> BCD is isosceles
Hence 45+x+45+x=150=> x=30
3rd Solution
ReplyDeleteLet the perpendicular bisector of AC, that is DE meet AB extended at E
Then B is an excentre of Tr. CDE and so < ADB = and hence x = 30
Sumith Peiris
Moratuwa
Sri Lanka
Last line should read
ReplyDeleteThen B is an excentre of Tr. CDE and so < ADB = 45 and hence x = 30
<ABD=135-x
ReplyDeleteConsider triangle ABD
sin(135-x)/AD=sin30/BD
AD/BD=2sin(135-x)---------(1)
Consider triangle CBD
sin15/BD=sinx/CD
CD/BD=sinx/sin15---------(2)
Since AD=CD, by equating (1) & (2)
2sin(135-x)=sinx/sin15
2sin(135-x)sin15=sinx
cos(120-x)-cos(150-x)=sinx
cos120cosx+sin120sinx-cos150cosx-sin150sinx=sinx
-cosx/2+sqrt(3)sinx/2+sqrt(3)cosx/2-sinx/2=sinx
(sqrt(3)-1)(sinx+cosx)=2sinx
(sqrt(3)-1)(1+cotx)=2
1+cotx=sqrt(3)+1
cotx=sqrt(3)
x=30