Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.
Click the figure below to view more details of problem 1211.
Monday, May 9, 2016
Geometry Problem 1211: Right Triangle, Altitude, Angle Bisector, 45 Degrees
Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.
Click the figure below to view more details of problem 1211.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.
Click the figure below to view more details of problem 1211.
Problem 1211
ReplyDeleteIs <EDB+<BDF=45+45=90, <EBF+<EDF=90+90=180,then B,E,D,F are concyclic.Then BE=BF=x
(<BEF=<BDF=45=<BFE).Therefore EF=x√2 . By the theorem of Ptolemy is BD.EF=DF.x+DE.x=
X(DF+DE) or DF+DE=(BD.EF)/x=BD. √2.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORRYDALLOS GREECE
http://s32.postimg.org/eng98hm51/pro_1121.png
ReplyDeleteLet G an are the projection of E over BD and F over DC ( see sketch)
Triangle BED similar to tri. CFD ( case AA)
So EG/FH= BD/DC= FH/HC= EG/BG => BG= FH
So EG+ FH= DG+BG= BD
Replace EG= ED. sqrt(2) and BG=FH=DF.sqrt(2)
We get ED+DF= DG+FH=BD.sqrt(2)
Let L be the line through B parallel to AC.
ReplyDeleteExtend DF to meet L at G.
Triangle BDG is right angled and isosceles (BD = BG)
So DG = BD√2.
It suffices to prove DE = FG.
Compare triangles BED and BFG.
Angle EBD = Angle FBG (each = C).
Angle EDB = Angle BGF (each = 45 degrees).
Side BD = Side BG.
By ASA the above triangles are congruent.
So DE = FG as required.
N Vijaya Prasad,
Rajahmundry,
INDIA.
In Quadrilateral BEDF, B+D = 90 and E+F = 135-c+45-c = 90 degrees. So it is cyclic.
ReplyDeleteJoin EF, since B,E,D, and F are concyclic, BDF = 45 AND FEB =45 degrees. Hence triangle BEF is right isosceles triangle => BE=BF=(1/Sqrt 2)EF -----> (1)
Since triangle DEF is right angle => Square(BE)=0.5(Square ED + Square FD) -----> (2)
Triangles EDB & FDC are similar
=> DC = BD(FD/ED) ----- > (3)
& CF = BE(FD/ED) -------> (4)
Triangles AED & BFD are similar
=> AD = BD(ED/FD) -------> (5)
& AE = BF(ED/FD) = BE(ED/FD) (From (1)) --------> (6)
From (3) & (5) AC = AD+DC = BD.(Square ED + Square FD)/(ED.FD) --------> (7)
From (1) & (4) BC = BF+CF = BE + BE(FD/ED) = BE (FD+ED)/FD ---------> (8)
From (1) & (6) AB = BE + AE = BE(FD+ED)/FD -------> (9)
Applying Pythogrous to Triangle ABC => Square AC = Square AB + Square BC
=> Square BD. Square(Square ED + Square FD)/Square(ED.FD) = Square(BE).Square(FD+ED)/Square(FD) + Square(BE).Square(FD+ED)/Square(ED)
Replacing BE = 0.5(Square ED + Square FD) from (2) in above equation and solving
=> FD + ED = Sqrt(2) BD
∠ADE+∠EDB=∠ADB
ReplyDelete2∠EDB=90
∠EDB=45
∠BDF=45
∠EDF=∠EDB+∠BDF=45+45=90
∠EBF+∠EDF=90+90=180
E,B,F,D concyclic
∠BEF=∠BDF=45
∠BEF+∠EFB+∠FBE=180
45+∠EFB+90=180
∠EFB=45
∠BEF=∠EFB
BE=BF
∠ADB=∠ABC=90
∠BAD=∠CAB
BAD~CAB
BD/DA=AB/BC
AE/EB=c/a
EA=cBE/a
BE+EA=BA
BE+cBE/a=c
aBE+cBE=ac
BE=ac/(a+c)
BF=BE
EF^2=BE^2+BF^2=2BE^2=2a^2
BD/BA=CB/AC
BD/c=a/√(a^2+c^2)
BD=ac/√(a^2+c^2)
c^2/(a+c)^2
(ED+DF)^2-EF^2=2BD^2-2a^2c^2/(a+c)^2
ED^2+2EDDF+DF^2-EF^2=2a^2c^2/
(a^2+c^2)-2a^2c^2/(a+c)^2
2EDDF=[(a+c)^2—(a^2+c^2)]2a^2c^2/(a^2+c^2)(a+c)^2=2ac2a^2c^2/((a^2+c^2)(a+c)^2)=4a^3c^3/((a^2+c^2)(a+c)^2)
Area of DEF=1/2EDDF=1/4(2EDDF)=1/4(4a^3c^3/(a^2+c^2)(a+c)^2)=a^3c^3/((a^2+c^2)(a+c)^2)
For ED+DF=sqrt(2)BD
ReplyDeleteLet P be the point where the lines EF and BD intersect.
Similarity of triangles BFD and EPD give :
(1.) BF/BD=EP/ED -------> BF*ED=BD*EP
Similarity of triangles BED and FPD give :
(2.) BE/BD=FP/DF -------> BE*DF=BD*FP
Adding equations 1 and 2 and using the fact that BF=BE=EF/sqrt(2) and EP+FP =EF gives :
BF*ED+BE*DF=BD*EP+BD*FP
(ED+DF)EF/sqrt(2)=BD(EP+FP)
(ED+DF)=sqrt(2)BD
For EF = ac*sqrt(2)/(a+c)
By similarity of triangle EDF and ABC we can get :
(3.) DF/EF = a/AC ------> DF*AC =aEF
(4.) ED/EF = c/AC ------> ED*AC =cEF
And the similarity of triangles ABC and ADB :
(5.) a/AC = BD/c ------> ac =AC*BD
Adding equations 3 and 4 and using DF+ED =sqrt(2)BD gives:
AC(DF+ED)=(a+c)EF
AC(sqrt(2)BD)=(a+c)EF
And using equation 5 gives :
ac*sqrt(2)=(a+c)EF
For Area DEF multiply equations 1 and 2 with eachother and use the previous results
(DE*AC)(DF*AC)=(ac)EF^2
AC^2(DE)(DF)=(ac)EF^2
AC^2(DE)(DF)=2(ac)^3/(a+c)^2
And AC^2 =a^2+c^2 thus
DE*EF = 2(ac)^3/(a^2+c^2)(a+c)^2
Dividing by 2 gives the area
I noticed the last part contains a few errors. I have therefore copied that part here and removed the errors
DeleteFor Area DEF multiply equations 3 and 4 with eachother and use the previous results
(DE*AC)(DF*AC)=(ac)EF^2
AC^2(DE)(DF)=(ac)EF^2
AC^2(DE)(DF)=2(ac)^3/(a+c)^2
And AC^2 =a^2+c^2 thus
DE*DF = 2(ac)^3/(a^2+c^2)(a+c)^2
Dividing by 2 gives the area