Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 4Triangle, Quadrilateral, Angles, Congruence. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
Trazamos el segmento BD El triangulo BDC es isosceles => angulo(DBC)=angulo(BDC)=(180-2A)/2= 90-ATomamos AD=DC=BC=1 En el triangulo BDC ; 1/sen(90-A)=BD/sen(2A) => BD=2 sen(A)En el tringulo ABD ; 1/sen(ABD)=2sen(A) / sen (A) => sen(ABD)= 1/2 => angulo (ABD) =30Angulo(x)=angulo(DBC)+angulo(ABD)=90-A + 30= 120 - A
It is possible to provide a proof using only elementary geometry!
Let point E be a reflection of point D over AB.Then AEB and ADB triangles are congruent Angle EAB = angle DAB = alpha => angle EAD = 2 alpha => triangles AED and CBD are congruent (AE = AD = DC = BC) => ED = DBBut BE = DB (because AEB and ADB are congruent) => EBD is an equilateral triangle.Angle EBD = 60 => angle ABD = 30 (since ABD = ABE)Angle DBC = 90 - alpha (since DCB is an isosceles triangle) x = angle ABD + angle DBC = 30 + (90 - alpha) = 120 - alpha.
This comment has been removed by a blog administrator.
http://mate-facil.co.cc/35/ By TiNo
Let the bisector of <DCB intersect AB at X. Now since <XCA = <XAD = a, AD = CD and XD is common, by SAS, AXD is congruent to XCD. Hence <DXC = <DXA. Since X is on the perpendicular bisector of BD it also follows that <DXC = <BXC. Therefore since <DXC = <DXA = <BXC and all three angles are supplementary, <BXC = 60. Hence <XBD = 30. Now, <XBC = <XBD + <DBC = 30 + 90 - a = 120 - a.
Video solution (Spanish version) by Eder Contreras and Cristian Baeza at Geometry problem 4. Thanks Eder.
solution is uploaded to the following linkhttps://docs.google.com/open?id=0B6XXCq92fLJJajJOTFRfNTlSTDZGdUlVZEJfSml4UQ
Proof:Join ACThrough D draw a line perpendicular to AC and meet AC at M and meet AB at E.∵ AD = DC∴ EM is the perpendicular bisector of AC∴ AE = EC and ∠AEM = ∠CEM∴ ∠EAD = ∠ECD = α∵ DC = CB ∠DCE = ∠BCE = α Δ DCE ≡ Δ BCE ∠CEM = ∠CEB∵ ∠AEM = ∠CEM = ∠CEB = 60o∵ In ΔBCE, x + α = 120o
Join BD.Let E be the circumcentre of ∆ABD.∠DEB = 2∠DAB = 2α.∆s DEB, DCB are congruent (Each isosceles, common base BD, equal vertical angles, each 2α)Follows EA = ED = DC = DA, ∆EAD is equilateral, and ∠AED = 60°So ∠ABD = (1/2)∠AED = 30°Observe that ∠DBC = 90° - αHence x = 30° + (90° - α) = 120° - α
hallo my dear brothers and sisters ...... here is a simple solution for this problemlets startjoin A,C. join B,D.draw an angle bisector of angle DCB.it intersects DB at the point F andalso intersects AB , at the point E.now consider triangle DBCCD=CB (given)angle DCF=angle FCB=alphaangle FDC = angle FBC = (180- 2 alpha)/2=90-alphaso angle DFC = angle BFC = 90so angle EFD = angle EFB = 90now consider two triangle DFC and BFCcf is common angle DCF=angle FCB=alphaangle DCF=angle FCB=alphatriangle CDF is congruent with triangle CBFso FD = FBnow consider two triangle DEF and EFBEF is common FD = FBangle EFD = angle EFB = 90two triangle DEF and EFB are congruent with each otherso angle BEF = angle FED............... (1)consider triangle ADCAD = DCthis is an isosceles triangleso angle DAC = angle DCAconsider triangle AECangle EAC = angle EAD + angle DACangle ECA = angle ECD + angle DCAangle EAD = angle ECD = alphaso angle EAC = angle ECAso triangle AEC is isosceles triangleso AE = ECnow consider two triangle AED and CEDAE=CEED is commonAD=DCtriangle AED and CED are congruentso angle AED = angle FED ..........(2)from (1) and (2)angle BEF = angle FED = angle AED = 180/3 =60so angle EBF = 90 - 60 = 30ANGLE ABC= angle EBF + angle FBC= 30 +90 - alpha = 120 - alpha
Draw an angle bisector of angle BCD and let it meet at the point E on the side ABThen connect E and D.Since In Tri ACD , angle ACD = angle CAD angle ECD = angle EAD = alpha CE = AE ( Isosceles tri AEC )Since In Tri BCE and Tri ADE BC = AD ( given ) angle BCE = angle EAD = alpha CE = AE (proved) Tri BCE congruent to Tri ADE. This implies x = angle CBE = angle ADESince in Tri CDE and in Tri ADE CE = AE ( proved ) CD = AD ( given ) angle ECD = angle EAD Tri CED congruent Tri ADE , angle CDE = angle ADE = xSince Tri BCE , Tri CDE , Tri ADE are congruent to each other angle BEC = angle CED = angle AED and BEC + CED + AED =180 degree 3(AED) = 180 degree AED = 60 degreeIN Tri CEA , angle CEA = 2(AED) angle = 120 degree. Since it Is isosceles angle ECA = angle EAC = 60 degreeIN Tri CDA , angle DCA = angle DAC = 60 - alpha , and angle CDA = 120 + 2(alpha)AND FINALLY ANGLE CDA = 360 - (ANGLE EDC + ANGLE EDA) ANGLE CDA = 360 - (ANGLE EDC + ANGLE EDC) ( EDC = EDA) ANGLE CDA = 360 - ANGLE 2(EDC) ANGLE CDA = 360 - 2X (EDC = X) 120 + 2(ALPHA) = 360 - 2X (ANGLE CDA = 120 + 2(ALPHA)) 2X - 360 = - 2(ALPHA) - 120 2X = 360 - 120 - 2(ALPHA) = 240 - 2(ALPHA) = 2( 120 - ALPHA) THEREFORE X = 120 - ALPHA
Let the bisector of < DCB meet AB at E and BD at F. Draw altitude DG of Tr. ABD.Tr. s BCF & DCF are congruent and these are also congruent with Tr. ADG. So DG = DF = FB hence in right Tr. BGD BD = 2DG which means that < ABD = 30 and since < DBC = 90-@ the result followsSumith PeirisMoratuwaSri Lanka