Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below for more details.
By problem 778, BH² + AC² = 4R² or R² + AC² = 4R² which gives AC =√3*R. Moreover, AC/sin(B) =2R or sin(B)=AC/(2R) =√3*R/(2R)=√3/2 which implies that /_B = 60° ------(1)Now refer to problem 1062. OH is the Euler line of a triangle with vertex /_B = 60°. Hence, CE-AD=OH, BC-AB=OH. This further means, CE - AD = BC - AB or AB - AD = BC - CE or BD = BE --------(2)(1) & (2) together imply that Tr. BDE is equilateral.
Let AO meet the circumcircle at N. Now <HCB =90-B= <NCB and hence HC//BN. But BH//NC, so BHCN is a //ogram and BH = NC = BO =ON =OC. Hence Tr. ONC is equilateral and so <B = 60.So < AHC =120 = <AOC and hence AHOC is cyclic. Also < CAO =30 =<CHO. Hence <BDE =60. So Tr. BDE has 2 angles = to 60 and the same is therefore equilateral.Sumith PeirisMoratuwaSri Lanka
Pl draw a figure if necessaryIt is well-known that BH = 2.OX where OX is perp bisector of ACand that OX = R cos BSo 2R cos B = BH = R implies B = 60 degDrop BF perpendicular to HO and extend it to meet the circle O at K.Easy to see that BHKO is a rhombusEqual arcs AK, CK subtend equal angles at B (each 30 deg)BK perp to DE implies each of the angles BDE, BED = 60 degHence triangle BDE is equilateral