## Sunday, May 18, 2008

### Elearn Geometry Problem 4 See complete Problem 4
Triangle, Quadrilateral, Angles, Congruence. Level: High School, SAT Prep, College geometry

1. Trazamos el segmento BD

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

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

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

3. 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 = 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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5. http://mate-facil.co.cc/35/
By TiNo

6. 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.

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

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

9. 10. Proof:
Join AC
Through D draw a line perpendicular to AC and meet AC at M and meet AB at E.
∴ 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

11. 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° - α
Hence x = 30° + (90° - α) = 120° - α

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

13. Draw an angle bisector of angle BCD and let it meet at the point E on the side AB
Then 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 )
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

14. 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 follows

Sumith Peiris
Moratuwa
Sri Lanka

15. 16. Construct point E such that E lies on AB and EC is an angle bisector of <DCB. Hence, <DCE=<BCE.
Construct line ED.
Triangle DCE is congruent to BCE. Hence <EDC=x