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

Labels:
angle bisector,
congruence,
isosceles,
median,
midpoint,
parallel,
triangle

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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