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

Labels:
congruence,
isosceles,
midpoint,
transversal,
triangle

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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.