## Thursday, July 8, 2010

### Problem 469: Isosceles triangle, Midpoint, Transversal, Congruence

Geometry Problem
Click the figure below to see the complete problem 469 about Isosceles triangle, Midpoint, Transversal, Congruence.

Complete Problem 469

Level: High School, SAT Prep, College geometry

1. 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
Ajit

2. Draw perpendicular lines DG and FH to line AC.
Two 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.

3. Consider triangle FDB &transversal ACE

1.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)

Q.E.D

4. By Menelaus AB/AD*DE/EF*CF/BC=CF/AD=1 because AB=BC and DE=EF.

Ivan Bazarov

5. Draw DG // AC where G is on CB extended

From mid point theorem CF = CG = CB + BG = AC + BD = AD since Tr. BDG is isoceles

Sumith Peiris
Moratuwa
Sri Lanka

6. Let x = AB = BC and y = BD (so AD = x + y).
1. 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.