## 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.

1. Problem 1211
Is <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

2. http://s32.postimg.org/eng98hm51/pro_1121.png

Let 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)

3. Let L be the line through B parallel to AC.
Extend 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.

Rajahmundry,
INDIA.

4. In Quadrilateral BEDF, B+D = 90 and E+F = 135-c+45-c = 90 degrees. So it is cyclic.

Join 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

2∠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
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)