## Sunday, July 13, 2014

### Geometry Problem 1029: Right Triangle, 90 Degrees, Angle, Midpoint, Congruence

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1029.

1. Extend DE to F so that ED=EF => BDCF is a parallelogram
Note that ∠(FDC)= ∠(DFB)= ∠(DBA) => BA is a tangent to circumcirle of triangle DBF
Circumcenter of DBF will be on BC => E is the circumcenter of DBF
So EB=ED=EC => BD perpendicular to AC

2. Is this solution is correct? :
1)Continue DE till point F such that BE = CE = EF --> BFC right triangle and E is the circumscribed circle
of BFC and since BC is the diameter And angle ABC is right so AB is tanget and so on ....

3. Hi Antonio

Please publish full of my problema.

Prob 1027
We build AH perpendicular BC and DF perpendicular AB
CH=HD=1/2DC=1/2ED
DF=DH=1/2ED
α =10
X=90-4α
X=50

Prob 1028

BE=ED=EC EM=EN EN =1/2 EC
Triangle ECN < ECN =30 ECN = 90-4X -> X =15

Prob 1029
We note <BDF = 90
BD perpendicular AC

Erina New Jersey

4. as stated, problem 1029 is wrong.

Given right ABC and BE=CE: ∠(DBA)=∠(EDC) does not imply BD⊥AC

Take D' on AC such as ED'⊥BC, then ∠(D'BA)=∠(ED'C) but BD' is not ⊥AC

bleaug

1. To Bleaug
Problem 1029 There is a condition
Thanks

2. naahhh, seriously?
then you should add this in the "Given:" section
bleaug

3. To Bleaug
Now it has been included in the "Given" section. Thanks

5. Thanks Antonio for the clarification.

Above solutions do not make explicit use of this newly added premiss, therefore they must be incomplete at the very least. Without it, there are two solutions to the problem, only one obeying CD>AD. For example in Peter's solution, there is no justification of the implication on the 3rd line.

Background: if we make D vary on AC and let F on ED be such that ∠(EDC) = x = ∠(FBA), it is easy to show that ∠(BFE) is constant and equal to ∠(BAC). See http://bleaug.free.fr/gogeometry/1029.png

Therefore, the locus of all F when D varies lies on the circle circumscribed to FEB. This circle crosses AC in at most two points. With simple geometric arguments we can show that the two following points belong to both AC and circle FEB and thus can be candidate solutions: D1 such as BD1⊥AC and D2 midpoint of AC. Since AD>CD, only D1 is acceptable => BD⊥AC

bleaug

6. Draw FE perpendicular to BC to meet BD extended at F.

Since AB // FE < EDC = < ABF = < BFE = < CFE since FE is the perpendicular bisector of BC. So CEDF is cyclic and BD is thus perpendicular to AC

Sumith Peiris
Moratuwa
Sri Lanka

7. Another method - let GE be the perpendicular bisector G on AC. Show that < CAB = ABG = < BGE = < CGE. Hence < DEG = < DBG , hence BDGE is cyclic and the result follows

Sumith Peiris
Moratuwa
Sri Lanka