Geometry Problem. Post your solution in the comment box below.

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

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## Monday, November 28, 2016

### Geometry Problem 1290 Triangle, Internal Angle Bisector, Median, Parallel, Measurement, Metric Relations

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https://goo.gl/photos/SxYFtYyAsxydH45r7

ReplyDeleteLet ME extended meet BC at F

Draw AN //ME . N is on BC extended ( see Sketch)

Observe that ∠BEF=∠BFE= u => triangle EBF is isosceles

∠BAN=∠ANB=u => triangle ABN is isoceles

Since M is the midpoint of AC => F is the midpoint of NC=> FN=FC

AB=BN=FN+FB=FC+x

BC=FC-FB =FC-x

So AB-BC= FC+x-(FC-x)=2x= 5 => x= 2.5

With the usual notation for triangle ABC,

ReplyDeleteDC = ab/(a+c) and so

MD = b/2 - ab/(a+c) = (b/2)(c-a)/(c+a)

Now ME//BD so

(c-x)/x = (b/2)/MD from which

c/x = 1 + (b/2)/MD

= 1 + (c+a)/(c-a) = 2c/(c-a)

Hence x = (c-a)/2 = 5/2 = 2.5

Sumith Peiris

Moratuwa

Sri Lanka

Extend AB to point F, such that BF=BC

ReplyDeleteAngle AFC= Angle ABC/2, hence FC is parallel to ME.

Since M is midpoint of AC,E must be midpoint of AF, AE=(AB+BC)/2. BE=AB-AE=(AB-BC)/2

BE=5/2

Problem 1290

ReplyDeleteDraw AP //BD//ME . P is on AB extended.Is <BCP=<DBC=<DBA=<BPC, then

BP=BC and AM=BC so AE=EP or AB-BE=BD+BE or AB-BC=2.BE or x=2.5.

MANOLOUDIS APOSTOLIS4 HIGH SCHOOL OF KORYDALLOS PIRAEUS-GREECE

Extend ME to cut BC at F, construct the parallelogram CFEG; due to AM=CM, ME passes through the midpoint of AG and, as it is the angle bisector of <AEG, tr. AEG is isosceles and AE=GE(=CF). With BF=BE, we are easily done.

ReplyDeleteLet AB = c, BC = a and AC = b

ReplyDeletewe have DC = ab/a+c , AD = bc/a+c and AM = b/2

Since triangles AME and ADB are similar

AE/AM = AB/AD

=> c-x/(b/2) = c/(bc/a+c)

=> 2c-2x = a+c

=> x = c-a/2 = 2.5

Draw CF Perpendic to BD, MN //CF

ReplyDelete=> MN = 1/2 AE = 2.5 (MG middle line of AFC)

=> MG=EB (MGBE parallelogram)

To c.t.e.o

DeleteWhere are points F and G?

MN is not necessary, use MG. Thanks

G point CF meet BD, F on AB (Tr BFC isoceles)

ReplyDeleteMN is also middle line of AFC, a way to prove equality of tr MNE and DCB

Thanks