## Friday, November 13, 2009

### Problem 385. Triangle, Angle bisector, Perpendicular, Parallel, Semiperimeter

Proposed Problem
Click the figure below to see the complete problem 385.

Complete Problem 385
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

2. Draw FG the altitude of Tr. BFC. Since BF bisects < B, < FBD =90 - A/2 - B/2 = C/2 = < FED since BEFD is cyclic and is also = < ECA. Hence DE is parallel to AC

So < BED = 90 - C/2 = < EBG considering Tr. BEC. Since BEDG is cyclic it follows that the chords subtending these = angles are also =. Hence DE = BG = s-b

Sumith Peiris
Moratuwa
Sri Lanka

3. Part 1:

Extend BE to meet AC at F and BD to meet AC at G.
AD is the perpendicular bisector of tr. ABG so BD = DG and likewise
BE is the perpendicular bisector of tr. CBF so BE = EF.
That means triangle BFG is similar to tr. BDE by SAS (in a 1:2 ratio)
and DE is parallel to AC.

Part 2:
DE = 1/2 FG from the ratio of the triangles
FG = AC - (AF + GC)
From the perpendicular bisectors AB = AG and BC = CF
Putting that together:
DE = 1/2 ( AC - ([AC - BC] + [AC - AB]) = 1/2(b -([b-a] +[b-c])
= 1/2(a + c - b) = 1/2 (a + b + c - 2b) = s - b

4. https://photos.app.goo.gl/QwRcGSzYzprO6DOx2

Let BD and BE meet AC at G and F
Note that triangles BAG and BCF are isosceles
And BD=DG , BE=EF
AG=AB=c and CF=CB= a
AF= b-a and CG= b-c
So DE//AC and DE= ½ FG= ½(b-(b-a)-(b-c))= ½(a-b+c)