See complete Problem 4

Triangle, Quadrilateral, Angles, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Sunday, May 18, 2008

### Elearn Geometry Problem 4

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angle,
congruence,
Problem 4,
quadrilateral,
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Trazamos el segmento BD

ReplyDeleteEl triangulo BDC es isosceles => angulo(DBC)=angulo(BDC)=(180-2A)/2= 90-A

Tomamos 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) =30

Angulo(x)=angulo(DBC)+angulo(ABD)=90-A + 30= 120 - A

It is possible to provide a proof using only elementary geometry!

ReplyDeleteLet point E be a reflection of point D over AB.

ReplyDeleteThen 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 = DB

But 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.

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ReplyDeletehttp://mate-facil.co.cc/35/

ReplyDeleteBy 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.

ReplyDeletehttp://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

ReplyDeleteVideo solution (Spanish version) by Eder Contreras and Cristian Baeza at Geometry problem 4. Thanks Eder.

ReplyDeletesolution is uploaded to the following link

ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJajJOTFRfNTlSTDZGdUlVZEJfSml4UQ

Proof:

ReplyDeleteJoin AC

Through 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.

ReplyDeleteLet 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° - α

Hence x = 30° + (90° - α) = 120° - α

hallo my dear brothers and sisters ...... here is a simple solution for this problem

ReplyDeletelets start

join A,C. join B,D.

draw an angle bisector of angle DCB.it intersects DB at the point F and

also intersects AB , at the point E.

now consider triangle DBC

CD=CB (given)

angle DCF=angle FCB=alpha

angle FDC = angle FBC = (180- 2 alpha)/2=90-alpha

so angle DFC = angle BFC = 90

so angle EFD = angle EFB = 90

now consider two triangle DFC and BFC

cf is common

angle DCF=angle FCB=alpha

angle DCF=angle FCB=alpha

triangle CDF is congruent with triangle CBF

so FD = FB

now consider two triangle DEF and EFB

EF is common

FD = FB

angle EFD = angle EFB = 90

two triangle DEF and EFB are congruent with each other

so angle BEF = angle FED............... (1)

consider triangle ADC

AD = DC

this is an isosceles triangle

so angle DAC = angle DCA

consider triangle AEC

angle EAC = angle EAD + angle DAC

angle ECA = angle ECD + angle DCA

angle EAD = angle ECD = alpha

so angle EAC = angle ECA

so triangle AEC is isosceles triangle

so AE = EC

now consider two triangle AED and CED

AE=CE

ED is common

AD=DC

triangle AED and CED are congruent

so angle AED = angle FED ..........(2)

from (1) and (2)

angle BEF = angle FED = angle AED = 180/3 =60

so angle EBF = 90 - 60 = 30

ANGLE 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 AB

ReplyDeleteThen 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 ADE

Since 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 = x

Since 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 degree

IN Tri CEA , angle CEA = 2(AED) angle = 120 degree. Since it Is isosceles

angle ECA = angle EAC = 60 degree

IN 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.

ReplyDeleteTr. 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 follows

Sumith Peiris

Moratuwa

Sri Lanka

https://www.youtube.com/watch?v=2XwhJStz4bA

ReplyDelete