Online Geometry theorems, problems, solutions, and related topics.
Proposed ProblemClick the figure below to see the complete problem 291.See also:Complete Problem 291Collection of Geometry ProblemsLevel: High School, SAT Prep, College geometry
In quad. DFBE we've DF & DE perpendicular BF & BE respectively. Hence DFBE is concyclic. Therefore, angle ABC = angle EDF = let's say, B. Now let angle EDB = B1 & angle BDF = B2. Hence, sin(B) = sin(B1+B2) = sinB1*cosB2 + oosB1*sinB2 = (BE/d)(DF/d) + (DE/d)(BF/d) = (BE*DF+DE*BF)/d^2However, by Ptolemy's Theorem, (BE*DF+DE*BF =d*eThus, sin(B)= d*e/d^2 = e/d ---------(1)Now area of Tr. ABC = acsin(B)/2. Hence, it's circumradius R = abc/4(acsin(B)/2) = b/2sin(B) or R = b/2(e/d) using equation (1). This gives:2R*e = b*dVihaan Uplenchwar, Dubai: email@example.com
Let M be the mid-point of BD & O the circumcircle of Tr. ABC. Then clearly, FM=DM=d/2 and /_FMB = 2*/_FDE=2*/_B since F,D,E & B are concyclic. Further, Tr. FME is isosceles. Likewise, Tr. OAC is isosceles with an apex /_ of 2*B. Hence, Tr. FME /// Tr. OAC which gives us, FM/FE = OA/AC or d/(2e) = R/b or R = bd/(2e)
Correction: Let O be the circum-centre of Tr. ABC.
Apply sine law in triangle ABC=> sin(B)= b/2RApply sine law in triangle EDF => sin(EDF)= e/dSince EDFB is cyclic quadrilateral so angle (B)= angle (EDF)So we have b/2R= e/d => 2R.e=bd