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.

## Monday, May 9, 2016

### Geometry Problem 1211: Right Triangle, Altitude, Angle Bisector, 45 Degrees

Labels:
45 degrees,
altitude,
bisector,
right triangle

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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)