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

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

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## Thursday, January 5, 2017

### Geometry Problem 1303: Triangle, 60 Degrees, Angle Bisector, Measurement

Labels:
60 degrees,
angle bisector,
measurement,
triangle

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Problem 1303

ReplyDeleteIf AD and CE intersect at I then <AIC=90+<ABC/2=120. So <DIC=60=<EBD, then the EBDI is cyclic, or <IDC=<IEB. Let point F on the side AC (no in extension) such that AE=AF.

Is triangle AEI=triangle AFI so <AEI=<AFI or <IEB=<IFC=<IDC so <FIC=<DIC and <DCI=<FCI.

Therefore triangle IDC=triangle IFC so CD=CF.Then AE+CD=AF+FC=AC.

APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

O, CE meet AD. Draw OK, OL OM, Perpendicular to tr sides

ReplyDeleteTr EOK = tr DOL,=> EK=DL => But AK=AM, CM=CL, =>

=> AK+CL=AM+MC => (AK-KE)+(CL+DL)=AM+MC

=>AE+CD=AM+MC => AE+CD=AC

Let AD, CE meet at X. Mark F on AC such that AE = AF.

ReplyDeleteTr.s AEX and AFX are congruent SAS and since < AXC = 120 (since < A + < C = 120),

< EXA = < FXA = < FXC = <DXC = 60.

Hence Tr.s DXC and FXC are congruent ASA.

So CD = CF

So AC = AF + FC = AE + CD

Sumith Peiris

Moratuwa

Sri Lanka

Let I the incenter, clearly DI=EI ( from cyclic BEID, with <EBI=<DBI=30) and as constructed, AF=AE.

ReplyDeleteIf F is reflection of E about AD, it lies onto AC and <AIF=<AIE=60, that is, it remains that <CIF=60=<CID; with EF=IE=ID we infer that triangles IFC and IDC are congruent (s.a.s.), thus FC=DC and we are done.

Best regards.

Solution 2

ReplyDeleteb^2 = a^2 + c^2 - ac ......(1) since < B = 60.

Now AE + CD = cb/(a+b) + ab/(b+c)

= b(bc+c^2+a^2+ab)/{(a+b)(b+c)}

= b(bc+b^2+ac+ab)/{(a+b)(b+c)}

= b

Sumith Peiris

Moratuwa

Sri Lanka

Solution 3

ReplyDeleteIf I is the incentre and r the inradius,

BI = 2r. Let ID = IE =p so DE=√3p.

Also a2 + c2 –ac = b2 since <B= 60

Applying Ptolemy to cyclic quadrilateral BDIE,

p.BE + p.BD = √3p.BI

So BE + BD = 2√3.r

c-AE + a-CD = 4√3.r = 4√3∆/s with the usual notation for area and semi-perimeter.

So AE +CD = a+c - 4√3.(1/2.ac.√3/2)/{(a+b+c)/2}

= a+c - 3ac/(a+b+c)

= (a2 + c2 –ac +ab+bc)/(a+b+c)

= (b2 + ab+bc)/(a+b+c)

= b

Sumith Peiris

Moratuwa

Sri Lanka