Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemClick the figure below to see the complete problem 511 about Triangle, Double Angle, Angle Bisector, Measure.
three bisectors meet on G, BE = c (E on BC)=> AGED rhombus => DEC isoceles => EC = ED = d=> BC = c + d
Draw CE perpendicular to BD and cut AB at F. ED cut BC at F.Since BD is angle bisector of angle (ABC) so triangle EBC is isosceles and BD become symmetrical line of triangle EBC.We have angle (BFD)=angle(BCA) =alpha (property of symmetric)And triangle EAD is isosceles. ( angle AED=angle ADE= alpha)AD=AE=BE-BA=BC-BAPeter Tran
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Another construction: Extend BA to E s.t. AE=AD or /_AED=/_ADE=α. Triangles EBD & CBD are ASA congruent or BE=BC or c+d=aAjit
INDSHAMAT from University of Kelaniya,Srilanka..Draw the angle bisector of AngleBAC which cuts BC at E.Then AE=EC(as angleEAC=angleECA)d/DC=c/a(BD is the Bisector)So DC=ad/c (1)angleAEB=2@ and because angleABC is common for both ABE & ABC triangles,ABC & ABE triangles are similar.Hence AE/(DC+d)=CE/(DC+d)=c/a=BE/c (2).So BE=c2/a,hence CE=a-BE=(a2-c2)/a (3).From(2)we have CE/(DC+d)=c/a.So applying values of(3)&(1)to CE&DC we have a2-c2=d(a+c).So (a-c)(a+c)=d(a+c).So d=a-c
hello , this is another way to find the result :A' in BC such that : BA = BA' . from here we have : BA'D = 2x and AD = A'D .and we have to triangle A'DC is isosceles (DA'C = 2x and C=x) .hence we are done because : A'C = A'D = AD .adil azrou ...
Extend CA to E such that Tr. ABE is isoceles. Then Tr. EBC is isoceles from which it follows that Tr. BED is isoceles. Hence a = d+c so d = a - c. Sumith PeirisMoratuwaSri Lanka
Problem 511On the side BC I take the point E such that AB=BE, then triangle ABD=triangle EBD so AD=DEand <DEB=<DAB=2a. But <BED=<EDC+<ECD, so <EDC=a. Then EC=ED=AD=d.Therefore d=EC=BC-BE=a-c.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE