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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 986
http://s25.postimg.org/7verkimrj/pro_986.pngWe have ∆ (BDC) similar to ∆ (ABC ) …. Case AASo ∠(ABC)=45Draw circumcircle of ADB, BC will tangent to this circle∠ (AOB)=2∠ (DBC)=90From D draw a line //OA . This line will bisect OB => ∠ (DOB)=60So x=∠ (DAB)= ½ ∠ (DOB)=30
Non-elementary proof: ΔABC~ΔBDCIf DC=1, AC=2, then BC=√2. By sine law, sin x / 1 = sin 45° / √2sin x = 1/2x = 30°
Circumcenter O of triangle DCB has <DOB=2(180-<ACB)=2(45+x)=90+2x, so <OBD=45-x=<ABD and O is on AB . x is always less than 45 so <ACB is obtuse so O is on the same side of the line DB as segment AB. P is on AC such that OP is perpendicular to DC. Therefore, CO^2=CP*CA=DC/2*DC*2=DC^2 because <COB=2*45=90, making triangle COD equilateral, 2x=60, x=30.
/_ABC = /_ABD + /_DBC = (45-x) + x = 45°. Draw CE perpendicular to AB. /_BCE = 45° and thus E lies on the perpendicular bisector of BC while /_CEB=2*/_CDB and thus E is the circumcentre of Tr. CDB which makes /_DEC = 2x and EC=ED. Now D is the circumcentre rt. Tr. AEC; hence ED=DC. In other words, Tr. EDC is equilateral or 2x=60° or x=30°
CH the altitude of Tr.BCD = 1/√2 ofCD. But BC=√2.CD from similar Tr.s. So BCH is a right Tr. with CH=1/2 BC. Hence x=30