Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemClick the figure below to see the complete problem 598.
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.
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° - 2αTwo angles in ∆BCE being α and 90° - 2α, the third angle ∠BCE = 90°+ αApplying Sine Rule in ∆BCE, BE : sin (90° + α ) = 2 : sin α3 sin α = 2 cos α, So tan α = 2/3 and AE = 3/sin α = 3 / [2 /√13] = 3√13 / 2From the right angled ∆ADE, tan α = x / AEHence x = AE tan α = (3√13/2).(2/3)(i.e.)x = √13
Let AB,DC meet at F. Triangles AFE and ADE are congruent ASA so EF = xNow BC = BE = 2 hence applying Pythagoras to triangle BEF, x = √(2^2 + 3^2) =√13Sumith PeirisMoratuwaSri Lanka