+5 As shown in the diagram, points B and D are on different sides of AC. We know that \(\measuredangle B = 2\measuredangle D = 60°\) and that \(AC = 4 \sqrt 3\). What is the distance between the circumcenters of \(\Delta ABC \) and \(\Delta ADC\)?
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Since Angle B = 2Angle D and they are both external angles of triangle ADC then Triangle BDC is isosceles. Let O be the circumcenter of triangle BDC. Since Triangle BDC is isosceles, BO = CO = r. Drop a perpendicular from B to AC at point E. Since BE = r, then BD = rsqrt(3). Also since Angle B = 60 then Triangle BDC is equilateral and therefore, Angle CBD = 120 degrees. Since Angle C = 90 - Angle B = 90 - 60 = 30 degrees then Angle ADC = 90 - Angle C = 60 degrees. Since Angle ADC = 60 degrees and BD = rsqrt(3), then AD = 2rsqrt(3). Now consider Triangle ADE. Since Angle ADE = 90 degrees and AE = 2r and AD = 2rsqrt(3), then DE = 2rsqrt(3). Also since DE = 2rsqrt(3), then CD = 2rsqrt(3). Since Angle C = 30 degrees and CD = 2rsqrt(3) and BD = rsqrt(3) then Angle BDC = 120 degrees. Therefore, Triangle BDC is equilateral and therefore, BD = r. Since BD = r and CD = 2r then by the Pythagorean Theorem, BC = rsqrt(3). Now consider Triangle ABC. Since Angle B = 60 degrees and BC = rsqrt(3) and AC = 4sqrt(3), then Angle A = 90 degrees. Therefore, Triangle ABC is a right triangle with legs AC = 4sqrt(3) and BC = rsqrt(3). Therefore, the circumradius of Triangle ABC is AC/2 = 2sqrt(3). Now consider Triangle ADC. Since Angle D = 30 degrees and CD = 2r and AC = 4sqrt(3), then Angle A = 60 degrees. Therefore, Triangle ADC is a right triangle with legs AC = 4sqrt(3) and CD = 2r. Therefore, the circumradius of Triangle ADC is AC/2 = 2sqrt(3). Therefore, the distance between the circumcenters of Triangle ABC and Triangle ADC are both 2sqrt(3) and therefore, the total distance between them is 22sqrt(3) = 4*sqrt(3). So the answer is 4*sqrt(3).
