Friday, October 22, 2010

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

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

Problem 531: Triangle, Angle Bisector, Midpoint, Parallel, Perimeter, Semiperimeter, Math Education
See also:
Complete Problem 531

Level: High School, SAT Prep, College geometry

8 comments:

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

    ReplyDelete
  2. Draw CG parallel to BD ( G on extension of AB)
    Triangle 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

    ReplyDelete
  3. draw EG//BC, EH//AB => ▲EGF isoceles
    => s = BC + DC + BF = AE + AF

    ReplyDelete
  4. Sorry, FG=FB+BG, So FG=FB+BC.
    s=AE+AF=CE+BF+BC.
    (Correct 1st comment)

    ReplyDelete
  5. To c.t.e.o

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

    ReplyDelete
  6. To Peter

    AFE =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

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

    s = 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

    ReplyDelete
  8. or we can use bisector theoreme :
    we 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 ^^

    ReplyDelete