Geometry Problem

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

Click the figure below to see the complete problem 775.

## Friday, June 29, 2012

### Problem 775: Triangle, Isogonal lines, Congruent angles, Perpendicular, Metric Relations

Labels:
angle,
congruence,
isogonal,
metric relations,
perpendicular,
triangle

Subscribe to:
Post Comments (Atom)

Angle EBM = Angle DBF

ReplyDeleteSo Rt triangles EMB and DFB are similar

∴ x : 8 = BE : BD

Next angle HBE = angle GBD

So Rt triangles HBE and GBD are similar

∴ BE : BD = EH : DG = 25 : 28

Follows x : 8 = 25 : 28

Hence x = 50/7

Triangle BEM and BFD are similar (AA)

ReplyDeleteTriangle BDG and BHE are similar (AA)

Hence DG/EH =BD/BE =ED/EM

28/25 =8/EM

So EM =50/7

Triangle BDG similar to ∆ BEH ….. ( Case AA)

ReplyDeleteSo BD/BE= 28/25

Triangle BFD similar to ∆ BME ….. ( case AA)

So DF/EM=8/x=BD/BE=28/25

So x=50/7

Let be P on BD such that BP=BE, and let be Q and N the perpendiculars from P to AB and BC respectively. Note that BPQ and BME are congruent, and that BPQ is similar with BDF and BNP is similar with BGB (AA). Hence x=50/7

ReplyDeleteGreetings go-solvers.

x/8 = BD/BE = 28/25 from which x = 200/28 = 50/7

ReplyDelete