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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 734 details.
http://img812.imageshack.us/img812/4533/problem734.pngDraw line NEL// BMC ( see sketch)Denote angles x, y, z, t as angles shown on the sketchSince M is the midpoint of BC => E is midpoint of NLSince 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)MN/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:Extend 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:For convenience denote AD=BD=d, ∠DBE=xNote ∠EDC=A=∠BDEDraw MN∥DE(N on AC)MN∥AB and MN=(1/2)ABDE/MN = AD/AN = 2d/bFollows DE = AD.MN/AN=AD.2MN/2AN=AD.AB/AC = dc/bAlso in ∆BDE, by Cosine RuleBE²=BD²+DE²-2BD.DE cos A=d²+(d²c²/b²)-2(d²c/b)cos A =[d²/b²][b²+c²-2bc.cos A]=[d²/b²].a²So BE = da/bBE:BD:DE=(da/b):d:(dc/b)=a:b:cThus ∆DEB ∼ ∆ABCHence ∠DBE = ∠ACB
We know that ∠BAD = ∠BDE.Hence 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)=1Also ∆ABH ~ ∆HED, (HB/DH)=(AB/ED)Combining and we get (CA/AB)=(BD/ED).Q.E.D.
I used menelaus too....∵BE//AB∴∠ABD=∠BDE (alternate interior angles theorem)∵AD=BD∴∠ABD=∠BAD (base angles theorem)∴∠BDE=∠BADGoal: Prove that △ABC∼△DEB so that ∠DBE=∠ACBAM 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=1AC/AD=BK/DKAD/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 MX = c/2If AD = p = BD and DE = q(b/2)/p = (c/2)/q so p/q = b/cSo triangles ABC and DBE are similar and so the result followsSumith PeirisMoratuwa Sri Lanka