## Monday, January 20, 2014

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

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.

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

2. Since it is a 30-60-90 triangle, the length ratio would be 1:sqrt3:2
the remaining can be done by angle bisector length theorem

3. 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

4. 2 ways of doing it.

Draw 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

5. 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.

1. APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORRYDALLOS GREECE
Solution 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.