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 to enlarge it.
Let O is a circle through A, B and D where cuts AC and DC at E and F respectively.Let AD and BC intersect at GNote that A, B, E, F and D concyclicBecause ∠GBA an exterior angle of triangle ABC then m ∠GBA = 52.5 degree so that m∠DBC = 105 degree. This implies ∠DBC = 30 degreeBecause BADF cyclic then m∠BAF = 30 degree so that m∠EAF = 7.5 degree = m∠EDF (A, B, E, F and D concyclic)Because ∠AED an exterior angle of triangle ECD then m ∠AED = (7.5 + 15) degree = 22.5 degree = x (From wisna email: email@example.com)
To AnonymousRefer to line 5 "so that m∠DBC = 105 degree"Please provide more explanation how do you get 105 degree.http://s17.postimg.org/mmbwojrcv/pro_1074.pngPeter Tran
To Peter TranI am sory because there was an error in my solution in line 5.Thanks for your corection. But I am sure that x = 22.5 degree.I still working to find that solution.
With out loss of generality assume AB = 1.(all angles are measured in degrees)First note CAD = 67.5,ADC = (180 - 82.5),ABC = (180 - 57.5) By Sine Rule applied to Δ ABC:AC / sin 57.5 = AB / sin 30 = 2.So AC = 2 sin 57.5 = 2 cos 37.5 ..... (i).By Sine Rule applied to ΔADC:AD / sin 15 = AC / sin 82.5 = AC / cos 7.5So AD = AC. [sin 15 / cos 7.5]= 2 [cos 37.5].[2 sin 7.5]= 2[2 cos 37.5sin 7.5] = 2[sin 45 – sin 30] = 2[1/(sqrt 2) - 1/2]= (sqrt 2) - 1 = tan 22.5 So tan x = AD / AB = (tan 22.5) / 1 = tan 22.5Hence x = 22.5 degrees
Let O be the center of a circle through B, D and C where cuts AC at P so we have triangle BPO equilateral (I) ( Because m∠BOP = 2m∠BCP=30º.Now have m∠BOD = 2.m∠BCD = 90º so #BODA is cyclic because m∠BAD + m∠BOD = 90º+90º= 180º then m∠BAO = m∠BDO = 45º but m∠BAO = m∠BAC+m∠CAO then m∠CAO = 22.5º (II). Circunference ABOD cuts AC at Q but AC bissects ∠BAO then we have BQ = QO then using (I), triangles BPQ and QPO are congruents by equal sides then m∠BPQ = 30º = m∠BAP + m∠ABP ( ∠BPQ external angle) .: m∠ABP = 30º - 22.5º = 7.5º (III) but m∠PBD = m∠PCD = 15º (inscritive angles), finaly using (III) we have m∠ABD = m∠ABP + m∠PBD = 7.5º + 15º .: m∠ABD = 22.5 º
Let O be the circumcentre of ABC and let CO and AD meet at E.< AOB = 60 so Tr. OAB is equilateral. Also < BOC = 45 so < AOE = 75. Hence <OAC = OCA = 37.5. Hence < DCE = 22.5 and < OAE = 30 so it follows that AO = AETherefore Tr. BAE is right isoceles and so < AEB = 45 = < DCB implying that BCED is cyclic. So < DBE = < DCE = 22.5 and so x = 22.5Sumith PeirisMoratuwaSri Lanka