Geometry Problem

Click the figure below to see the complete problem 531 about Triangle, Angle Bisector, Midpoint, Parallel, Perimeter, Semiperimeter.

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Complete Problem 531

Level: High School, SAT Prep, College geometry

## Friday, October 22, 2010

### Problem 531: Triangle, Angle Bisector, Midpoint, Parallel, Perimeter, Semiperimeter

Labels:
angle bisector,
midpoint,
parallel,
perimeter,
semiperimeter,
triangle

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Extend Ray RB and draw line from C parallel to EF (or DB) till they intersect at G.

ReplyDeleteLooking at Triangle ACG, we have EF parallel to CG, and given AE=CE, so AF=FG. But DB is also parallel to CG, so Angle ABD=Angle AGC=Angle DBC=Angle BCG=Angle a. Since Angle BCG=Angle AGC (BGC),BC=CG, that means because AF=FG, FG=FB+BG, then FG=FB+GB. So AF=BF+BG.

Therefore, AE=CE, thus s=AE+AF=CE+BF+BG.

Draw CG parallel to BD ( G on extension of AB)

ReplyDeleteTriangle BCG is isosceles ( angle BCG= angle BGC = alpha)

So AG=AB+ BG= AB+BC

Since E is the midpoint of AC so F is the midpoint of AG. ( EF // CG)

And AE+AF= 1/2AC + ½( AG) = ½( AC+AB+BC) =semi-perimeter of triangle ABC

CE+BC+BF= AB+BC+EC-(AE+AF) = semi-perimeter

Peter Tran

draw EG//BC, EH//AB => ▲EGF isoceles

ReplyDelete=> s = BC + DC + BF = AE + AF

Sorry, FG=FB+BG, So FG=FB+BC.

ReplyDeletes=AE+AF=CE+BF+BC.

(Correct 1st comment)

To c.t.e.o

ReplyDeleteIt is not clear how do you get "=> s = BC + DC + BF = AE + AF " from "draw EG//BC, EH//AB => ▲EGF isoceles" .

Please explain

To Peter

ReplyDeleteAFE =α coorr. ang and GE//BC (BC middle line)

=> EGF + B = 180 => EGF = 180 - 2α

from ▲EGF => GEF = α => EGF isoceles

=> EG = GF = BH = HC

s = AE + ( EG + AG ) = AE + AF ( AF = EG + AG =GF +AG)

s = AE + ( EG + AG ) = CE + ((EG + GF ) + BF)=CE+BC+BF

ABC of semiperimeter "s" with the angle bisector BD. If E is the midpoint of AC and EF is parallel to BD, prove that

ReplyDeletes = AF + AE = CE + BC + BF

Sides and Line Segments

AB = AF + BF (TOP FLANK)

BC = BC (RUDDER)

AC = AE + CE (SOUTHERN BORDER)

AE = CE (S. BORDER DIVIDED EQUAL PARTS)

CE = DE + DC (LEFT S. BORDER HAS GULLY)

AE = DE + DC (EQUALITY REPEATED)

AD = AE + DE (SOUTHERN THRUWAY A --E -- D)

DC = CE - DE (REMOTE PART OF LEFT S. BORDER)

DE = CE - DC (GULLY)

PROOF

(i) E is the midpoint of AC, AC = CE, and a proof that s = AF + AE = CE + BC + BF

is equivalent to a proof that

AF = BC + BF or BC = AF - BF

(ii) BD bisects angle ABC. So,

\frac{BD}{DC} = \frac{AB}{AD}

Define BC as

BC = DC * \frac{AB}{AD} (Eq. 1)

(iii) EF is parallel to BD. So,

angle ADB = angle AEF.

Triangles AFE and ABD share angle CAB.

Thus, triangles AFE and ABD have

two angles with identical measurements.

Since, the sum of angles in a triangle

is 180, the third angle in the two

triangles are equal.

So, triangles AFE and ABD are similar

and have proportional corresponding sides.

\frac{AF}{BF} = \frac{AE}{DE} (Eq. 2}

Further,

DE = AE * \frac{BF}{AF}

(iv) Rewrite Eq. 1

BC = DC *\frac{AB}{AD}

to contain only the variables that we need in the proof (AF, BF and BC).

(a) DC = CE - DE

DC = CE - \frac{AE * BF}{AF}

(b) AB = AF + BF

(c) AD = AE + DE

AD = AE + \frac{AE * BF}{AF}

Thus Eq. 1 becomes,

BC = {CE - {AE * BF}{AB}} * \frac{AF + BF}{AE + \frac{AE * BF}{AF}}

(Eq. 3)

Multiplying both sides by

(AE + \frac{AE * BF}{AF})

to obtain

BC * (AE + \frac{AE * BF}{AF}) =

(CE - \frac{AE * BF}{AF}) * (AF + BF)

wich furer simplifies to

(AE*AF*BC)+ (AE*BC*BF) = (AF*AF*CE) + (AF*BF*CE)- (AE*AF*BF) - (AE*BF*BF)

Factoring, dividing all terms by AE=CE derives

(AF*BC) + (BC*BF) = (AF*AF) + (AF*BF) - (AF*BF) - (BF*BF)

(Eq. 4)

(v) The right side of the equation contains the exact expansion of AF^2 - BF^2. So rewriting Eq.4 yields,

BC*(AF + BF) = (AF+BF)*(AF-BF)

-- or --

BC = AF - BF

or we can use bisector theoreme :

ReplyDeletewe have : DC\AD = BC\AB <=> AC\AD = (BC+AB)\AB

and we have by thales : 2AF\AB = AC\BD = (BC+AB)\AB => 2AF = BC + AB ... and we are done !!

By adil azrou ^^