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
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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 ^^