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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 view more details of problem 1173.
Consider Tr.DBA and Tr. ABC, AB/BC = DB/AB = 2/3 Angle B is common, thus Tr.DBA and Tr. ABC are similar hence Angle BAD = Angle CAlso AD/AC = AD/(AE+CE)=2/3, Since AD=AE we get AD=2CELets assume bisector of angle B meets AC at F, AF/FC=2/3. FC=3AC/5 And CE=AC/3 and it is easy to see FE/CE = 4/5, hence BF//DE and it proves Angle ABD = 2 × Angle CDE.
TrekABDingjashem me trekABC (BKB) kendBAD= kendACB=C kendADB=kendBAC=Ashenoj kendCDE=a kendADE=kendAED=x x=c+a kendADC=x+a=c+2a kendADC=kendABC+C kendABC=2a=2CDE 3AD=2AC AD=2AC-2AD=2EC
http://s15.postimg.org/o5x05aenv/pro_1173.pngVerify that BA^2=BD.BC => BA tangent to circumcircle of tri. ADCSo ∠(BAD)= ∠(ACD)Draw BF//DEIn triangle ABF we have ∠(ABF)= ∠(BFD)- ∠ (BAD)In triangle DEC we have ∠ (EDC)= ∠ (AED)- ∠ (ACD)But ∠ (AED)= ∠ (ADE)= ∠ (BFD)And ∠ (FBD)= ∠ (EDC)Replace in above expressions we get ∠ (ABD)=2. ∠ (EDC)Triangles DEC and BFA are similar ..( case AA)So CE/AF=DC/AB= 5/6Since BF bisect angle ABD soAF/AB=FD/BD= > AF= 3/5 x ADAnd AD=AE=2 x CE
In Triangles BAD, BCA we have BA/BC=6/9=2/3=4/6=BD/BA and angle B is common. By SAS they are ///.So AE/AC=AD/AC=2/3 and thus EC/AC=1/3, AE/EC=2/1, AD=AE=2EC.Also Corresponding anglea BAD, BCA are equal. Hence Angles ABD=ADC-BAD=ADE+EDC-BAD =(AED-BAD)+EDC=(AED-C)+EDC=CDE+CDE=2CDE
AB/BC=6/9=4/6=BD/BA ,therefore tri ABD and tri CBA is similar, then <BAD=<BCA.The projections of ED intersects AB in K . Is <BAD=<ACB=x and <EDC=y, then <ADE=<AED=x+y, but <AKD=<ADE-<KAD=y.Therefore <ABC=<AKD+<KDB=2y=2<EDC.Bring from C parallel to the KE intersecting AK to L( CL//KE), then is KL=DC=5,BK=BD=4AK=10, but AE/EC=AK/KC=10/5=2.Therefore AE=2EC.
AB/BC=6/9=4/6=BD/BA ,therefore tri ABD and tri CBA is similar, then <BAD=<BCA.The ED intersects AB at F . Is <BAD=<ACB=x and <EDC=y, then <ADE=<AED=x+y, but <ADE=<FAD+<AFD=x+y,then <AFD=y=<BDF , therefore BF=BD=4 and <ABD=<BFD+<BDF=2y=2<EDC.But <BKC=<BFD=y=<BCKBring from C parallel to the DE intersecting AF to K( CK//FE), But <BKC=<BFD=y=<BCK then is KF=DC=5,BF=BD=4AF=10, but AD/EC=AE/EC=AF/FK=10/5=2.Therefore AD=2EC.
We use Barycentric Coordinates.Definition: Capital letters will denote vectors. P = x A + y B + z C will be written as P = (x, y, z). By definition, x + y + z = 1. AB means “length of AB” unless otherwise specified; most importantly, it does NOT mean “A times B”. Let A = (1, 0, 0), B = (0, 1, 0) and C = (0, 0, 1).As BD : DC = 4 : 5, it is clear that D = (0, 5/9, 4/9).If P – M = (u, v, w), then PM is given by:d^2 = -(vw)a^2 – (uw)b^2 – (uv)c^2, where a, b, c are BC, AC, AB respectively.Plugging in D – A into the distance formula results in AD = 2b/3. It is given that AE = AD, hence E = (1/3, 0, 2/3). It is obvious that (1/3)A + (2/3)C = E, hence AE:EC = 2:1, and therefore AD = 2CE. [One part of the question -> tick!]Construct G on such that G = (5/9, 0, 4/9). This means that AG : GC = 4 : 5, with G on line AC. This ratio and the fact that C is shared by both Triangle GDC and Triangle ABC indicate that the named triangles are similar and that AB is parallel to GD. From the similar triangles, AB : GD = BC : DC, so GD = 10/3. It’s trivial to find the following facts: E = (3/5)G + (2/5)C which indicates that EG : EC = 2:3; GD : DC = 2 : 3. By the angle bisector theorem, line DE bisects angle GDC. With parallel lines, Angle GDC = Angle ABC. Hence, twice of Angle EDC = Angle ABC. [Second part of question -> tick!]The third part of question is simple angle chasing.Angle AED = Angle CDE + Angle ECD. (External angle, internal opposite angles of Triangle CDE)Angle AED = Angle ADE (Isosceles triangle)Hence, Angle ADC = (Angle CDE + Angle ECD) + Angle CDEAngle ADC = Angle ABD + Angle BAD (External angle, internal opposite angles of Triangle ABD)Combine two expressions of Angle ADC.Angle ABD + Angle BAD = (Angle CDE + Angle ECD) + Angle CDETherefore, Angle BAD = Angle ECD. [Question done, boys!]