Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemClick the figure below to see the complete problem 599.
Drop BN perpendicular to ACw.l.o.g. let each side of the hexagon be 2Easy to see BN = 1, AN = √3, AC = 2√3 AM = AH/2 = (AC + CH)/2 = (2√3 + 2)/2 = √3 + 1NM = AM – AN = 1∆BNM is right angled isosceles,∠BMN = 45°x = 135°
it is a lot easier!!!! if you see <AHB=15° you can remember the special triangle 30-15-135.median traced to the largest side forma angle 45° and x=135°.
Let be O the center of square. Note that OM || AD and BC=OM. Thus triangles CMO and MCB are congruents. We know angle OCH = 45º, thus x = angle MCB=135º.
Problem 599Form the equilateral triangle BHK (points M and C are inside the triangle BHK).Then <BKH=60=2.30=2.<BAH so K is circumcenter of triangle ABH.So AK=BK=BH=HK and MK=MH=MA then triangle BMK=triangle BMH.Therefore < MBH=<MBK=30.Then <BMA=30+15=45. Therefore <BMH=135.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
Problem 599 solution 2Form the equilateral triangle BHK (points Β is inside the triangle BHK).Then Triangle ABK=triangle ABH then <AKB=<AHB=15 and BK=BH,<BKH=<BHK=45so <KBH=90=<KMH.Then KBMH is cyclic.So <MBH=<MKH=30.Therefore<BMH=135.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
Drop a perpendicular BP to ACH. Obviously PM= a/2 where a = side of theregular hexagon and the square. But BP also = a/2 considering the 30-60-90triangle BCP.Hence BP = PM and so x = 135Sumith PeirisMoratuwaSri Lanka