## Friday, April 29, 2011

### Geometry Problem 598: Trapezoid, Perpendicular, Equal Angles, Metric Relation

Geometry Problem
Click the figure below to see the complete problem 598.

1. Let H in AD that AEH is rectangle. Then AEB is congruent to AEH (angle ABE=90) and EH=3. Now BH is perpendicular to AE and thus is parallel to CD. Then HD=BC=2 and finally Pitagoras to EHD gives x=ED=sqrt(13).

MIGUE.

2. In ∆ABE, angle at A is α, angle at E is 90° - α. So angle at B is 90°
Since AD ∥ BC we have ∠ABC is 180° - 2α.
So ∠EBC is 90° - 2α
Two angles in ∆BCE being α and 90° - 2α,
the third angle ∠BCE = 90°+ α
Applying Sine Rule in ∆BCE,
BE : sin (90° + α ) = 2 : sin α
3 sin α = 2 cos α,
So tan α = 2/3 and
AE = 3/sin α = 3 / [2 /√13] = 3√13 / 2
From the right angled ∆ADE, tan α = x / AE
Hence x = AE tan α = (3√13/2).(2/3)
(i.e.)x = √13

3. Let AB,DC meet at F. Triangles AFE and ADE are congruent ASA so EF = x

Now BC = BE = 2 hence applying Pythagoras to triangle BEF,

x = √(2^2 + 3^2) =√13

Sumith Peiris
Moratuwa
Sri Lanka