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

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

This entry contributed by Mr Petre Ciobanu, Math Teacher from Sibiu, Romania.

Click the figure below to see the complete problem 963.

## Monday, January 20, 2014

### Geometry Problem 963: Right Triangle, 30-60-90 Degrees, Angle Bisectors, Metric Relations

Labels:
30-60,
angle bisector,
metric relations,
right triangle

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http://imagizer.imageshack.us/v2/800x600q90/203/r3th.png

ReplyDeleteMake AH ⊥BC and HF=HD (see sketch)

Triangle CAH is 30-60-90 triangle

Triangle AFD is isoceles

Triangles BEC similar to AFC .. ( case AA) => AF/BE=AC/BC= ½

Bu AF= AD => AD= ½ BE

Since it is a 30-60-90 triangle, the length ratio would be 1:sqrt3:2

ReplyDeletethe remaining can be done by angle bisector length theorem

Midpoint of EB is P. <DIB=60 where I is incenter, so CEPD is cyclic, so then since <ECI=<DCI=30, EI=ID. But <IAP=<IPA=30, so AI=IP. AI+ID=EI+IP=AD=EP=EB/2

ReplyDelete2 ways of doing it.

ReplyDeleteDraw altitude AH. Tr.s AHD and ABE are simiar from which the result follows

Or draw median AO of Right Tr. ABC. Tr.s AOD and BEC are similar from which the result follows.

Sumith Peiris

Moratuwa

Sri Lanka

F is a point on BC such that CA=CF. Then F is the mid point of BC. Triangles BCE and ADF are equiangular and similar as well. So, BE/AD=BC/AF=2/1. Therefore, AD=(1/2)BE.

ReplyDeleteAPOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORRYDALLOS GREECE

DeleteSolution 1

Even M is midpoint of BE.Bring EN perpedikular in BC.Then AM=NM=1/2.BE .Is AENB concyclic .Is <DAN=30, <AND=75 and <DNA=75, therefore AN=AD. Is <NMA=60 then triangle NMA is equilateral.

Therefore AD=AN=AM=1/2.BE.

Solution 2

EB intersects AD to I is incenter to triangle ABC .Then <DIB=60 =ACB therefore CEID is concyclic,

<BED=30=<ICD=<ICA=<ADE. The ED intersects AB to F. Is <EFA=15=<EBA , then AF=AB. Bring from B

Parallel to AD where intersects the ED to N. Is <BNE=<ADE=30=<BEN .So BE=BN. Is AD=1/2.BN=1/2.BE.