Geometry Problem
Click the figure below to see the complete problem 469 about Isosceles triangle, Midpoint, Transversal, Congruence.
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Complete Problem 469
Level: High School, SAT Prep, College geometry
Thursday, July 8, 2010
Problem 469: Isosceles triangle, Midpoint, Transversal, Congruence
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Consider triangle FBD with ACE as the transversal. By Menelaus's Theorem, (AB/AD)*(DE/DF)*(CF/BC)=1, ignoring the minus sign. Given DE=DF and AB=BC, we get CF/AD=1 or AD=CF
ReplyDeleteAjit
Draw perpendicular lines DG and FH to line AC.
ReplyDeleteTwo right triangle ADG and CFH are congruency, since DG = FH (since DE = EF),
and angle DAG = angle BCA = angle FCH.
So, we get AD=CF.
Consider triangle FDB &transversal ACE
ReplyDelete1.therefore by Menelau's thm.
AD/AB * BC/CF * FE/ED = 1
2.FE=DE.........(E midpoint)
therefore FE/DE=1
3.AD * 1/CF * 1 = 1.....(AB=BC...[given] &
from 1)
4.therefore AD=CF.......(rearranging)
Q.E.D
By Menelaus AB/AD*DE/EF*CF/BC=CF/AD=1 because AB=BC and DE=EF.
ReplyDeleteIvan Bazarov
Draw DG // AC where G is on CB extended
ReplyDeleteFrom mid point theorem CF = CG = CB + BG = AC + BD = AD since Tr. BDG is isoceles
Sumith Peiris
Moratuwa
Sri Lanka
Let x = AB = BC and y = BD (so AD = x + y).
ReplyDelete1. Draw EG // to AD where G is on BF. Since E is the midpoint of DF this creates a 1/2 scale version of BDF.
2. That implies EG is 1/2 BD = 1/2 y.
3. Angle chasing also shows CEG ~ ABC by AAA and therefore its isosceles and
EG = CG = 1/2 y.
4. Since tr EFG is a 1/2 scale version of BDF => FG = BG = x + 1/2 y.
5. Putting that together AD = AB + BD = x + y and CF = CG + FC = 1/2 y + (x + 1/2 y) which is the same.