Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the problem 734 details.
Wednesday, February 29, 2012
Problem 734: Triangle, Isosceles, Median, Parallel, Angle Bisector, Congruence, Midpoint
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http://img812.imageshack.us/img812/4533/problem734.png
ReplyDeleteDraw line NEL// BMC ( see sketch)
Denote angles x, y, z, t as angles shown on the sketch
Since M is the midpoint of BC => E is midpoint of NL
Since DE//AB and E is the midpoint of NL =>
DA=DL ( Thales’ theorem)
Angles x=y=z=t (alternate angles and angles of isosceles triangle)
Triangle DEB congruence to tri. DEL (case SAS)
So angle (DBE)= angle (DLE)= angle (DCB)
Draw MN ∥ ED (N on AC)
ReplyDeleteMN/DE = AN/AD = NC/BD and
∠MNC = ∠BAD = ∠ABD = ∠BDE
∴ ∆DBE ~ ∆NCM (s.a.s)
So ∠DBE=∠NCM=∠ACM
Here's a slightly different proof:
ReplyDeleteExtend BE to intersect DC in F. Consider tringle BFC with AEM as the transversal.
By Menelaus, (BE/EF)*(FA/AC)*(CM/MA)=1 ignoring the –ve sign. But CM=MA and BE/EF =BD/DF by the angle bisector theorem since DE bisects /_ADF; hence
BD/DF = AD/DF = AC/FA =(AD+DC)/(AD+DF) which gives AD^2=DF*DC or BD^2=DF*DC which implies that BD is tangent to the circumcircle of triangle BFC and thus /_DAE=/_C by the alternate segment theorem.
Ajit
Yet another solution to Problem 734:
ReplyDeleteFor convenience denote AD=BD=d, ∠DBE=x
Note ∠EDC=A=∠BDE
Draw MN∥DE(N on AC)
MN∥AB and MN=(1/2)AB
DE/MN = AD/AN = 2d/b
Follows DE = AD.MN/AN=AD.2MN/2AN=AD.AB/AC = dc/b
Also in ∆BDE, by Cosine Rule
BE²=BD²+DE²-2BD.DE cos A
=d²+(d²c²/b²)-2(d²c/b)cos A
=[d²/b²][b²+c²-2bc.cos A]=[d²/b²].a²
So BE = da/b
BE:BD:DE=(da/b):d:(dc/b)=a:b:c
Thus ∆DEB ∼ ∆ABC
Hence ∠DBE = ∠ACB
We know that ∠BAD = ∠BDE.
ReplyDeleteHence we just need to prove that 2 prop. sides for similar triangles to complete the entire proof.
By menelaus, |(BM/MC)*(CA/AD)*(DH/HB)|=1
But BM = MC , (CA/AD)*(DH/HB)=1
Also ∆ABH ~ ∆HED, (HB/DH)=(AB/ED)
Combining and we get (CA/AB)=(BD/ED).
Q.E.D.
I used menelaus too....
ReplyDelete∵BE//AB
∴∠ABD=∠BDE (alternate interior angles theorem)
∵AD=BD
∴∠ABD=∠BAD (base angles theorem)
∴∠BDE=∠BAD
Goal: Prove that △ABC∼△DEB so that ∠DBE=∠ACB
AM and BD intersect at K
∵△AKB∼△EKD
∴DE/AB=DK/BK
(Goal: Prove that DK/BK=BD/AC=AD/AC)
BM/CM*AC/AD*DK/BK=1 (Menelaus' theorem)
∵BM=CM (Def. of medians)
∴AC/AD*DK/BK=1
AC/AD=BK/DK
AD/AC=DK/BK=BD/AC
∴DE/AB=BD/AC
∵∠BDE=∠BAD
∴△ABC∼△DEB
∴∠DBE=∠ACB
If X is midpoint of AC, then MX//DE//AB and
ReplyDeleteMX = c/2
If AD = p = BD and DE = q
(b/2)/p = (c/2)/q so p/q = b/c
So triangles ABC and DBE are similar and so the result follows
Sumith Peiris
Moratuwa
Sri Lanka