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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
Fie BPperpendicular pe AC ,psituat pe AC,daca notam cu M intersectia dreptelor AC cu BP =>BMCE-patrulater insciptibil,AB=BC=a
Draw the circle with center B and radii BA . This circle is the locus of every point which see the segment AD with angle ACD therefor C lies on the circle, therefore CB = BA = a
Fie BP perpendicular pe AC ,P situat pe AC,daca notam cu M intersectia dreptelor AC cu BP =>BMDC-patrulater insciptibil, <BAM=<MDB=<ACB=.AB=BC=a
AD subtends an angle of 2α at B while it subtends an angle of α at D. Moreover, B lies on the perpendicular bisector of AD since AB=BD=a. Hence, B is the circumcentre of Tr. ADC whose circumradius is a= BA=BD=BC.