See complete Problem 8 at:

www.gogeometry.com/problem/problem008.htm

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

Post your solutions or ideas in the comments.

## Sunday, May 18, 2008

### Elearn Geometry Problem 8

Labels:
angle,
congruence,
triangle

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Angle ABC = 90, AB=BC =>ABC is isosceles right triangle => AB=BC=AC/sqrt(2)=>AD=AC/sqrt(2)

ReplyDeleteExtend DC line past point D. Let AE be a perpendicular that falls to the extension of DC.

Angle ADE = x + (45 - x) = 45. Since ADE is a right triangle, angle DAE=45=>AE=ED=AD/sqrt(2)=AC/2

AEC is a right triangle with AE = AC/2 => angle ACD=30=>x=15

after bilding AE perpendicular with CD , a circle

ReplyDeletecan prescribe ABCE with center O ( midle of AC )

Angle AOE = 2 ACE => angle AEO = EAO = 45 + x =>

AE = r = 1/2 AC => C=30 => x = 15

Aplicando la ley de Senos en el triĆ”ngulo ACD, pues AC=AD*sqrt(2). E·so por cuanto AB=BC=AD. Y listo, y el resto es ejecutar.

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

ReplyDeleteLet E be an intersection of the circumcircle of triangle ABC and the extension line of CD.

ReplyDeleteThen ang(ADE)45deg and AE=AC/2 and so AED is a triangle with ang(A)=60, ang(E)=90 and ang(C)=30. We get x=15degree.

Solution by Cristian

ReplyDeletehttp://www.youtube.com/watch?v=1eOjUdxIHso

Greetings :)

solution is uploaded to the following link:

ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJUi1TWldsSXhUODI5Zkl5bVNFNlREQQ

http://img9.imageshack.us/img9/3933/problem8.png

ReplyDeleteSince ∠ (CAD+ ∠ACD=45 so ∠ (ADC)=135

We have big angle ∠ (ABC)=360-90=270

Consider circle center B, radius BA=BC

Since ∠ (ADC)=1/2 center angle ∠ (ABC) => D will be on this circle ( see sketch)

And BA=BD=AD => triangle ABD is equilateral

X= ∠ (BAD)-45=15

In the usual notation c = a = AD and b² = 2a², b = a√2

ReplyDeleteDenote 45° - x = y

By Sine Rule

a/sin y = AD/sin y = AC/sin 135° = b√2

sin y = a/b√2 = 1/2,

45° - x = y = 30°,

x = 15°

draw perpendicular to EC from A,E is point on extension of CE

ReplyDeleteAC=a*sqrt2, angEDA=x+45x=45, angEAD=45, AD=a(given).so, AE=ED=a/sqrt2, AC=a*sqrt2, so,AC=2AE & angAED=90.so, angACE=30=45-x

x=15

To Sanoojan Baliah, problem 8.

ReplyDeleteI can't understand what point is E, defined in the sentence "draw perpendicular do EC from A, E is point on extension of CE".

Could you enlighten it to me?

Thanks.

E is on extension of CE,angel AED=90degree

DeleteTo Nilton Lapa

ReplyDeleteActually Sanoojan Baliah Says:

Produce CD and Draw a perpendicular to CD from A it is AE.

Greetings and THanks

R.PradheepKannah

To Rengaraj PradheepKannah.

ReplyDeleteThanks again for your help. Now I understand the solution presented by Sanoojan Baliah, it is the same idea of Ahmetelmas.

Solution- Solucao ( English- Portugues)

ReplyDeleteEnglish:

I am brazillian so i will try my best to write a solution in english.

Note that angle ACB= angle CAB= 45 deegres

so if we add a angle of X next to angle C we get an 90 degrees angle:

Lets place an triangle identical to ACD, where BC=AD and angle CAD meets angles ACD and ACD to form a right angle. Let J be the vertex of the triangle BCJ , where angles CJB=45-X , JCB= X and CBJ=135 degrees, let JB meet the line AC in P, hence the angle CBP=45 and ABP=45. This provides us with BPC=BPA=90 degrees.

Finally we can assume that AC equal 2L while AB=BC=AD= L times the square root of 2.

The sin law allow us to conclude that :

triangle ADC:

2L / sin 135 = L time square root of 2/ sin 45- x

Being X< 45 we conclude that

X=15 degrees.

Portugues:

Note que os angulos em A e C sao iguais a 45 graus.

Ainda, percebemos que os angulos em C sao ACB=45 e ACD= 45-X , logo se adicionarmos um angulo X teremos um angulo reto. Posicionando-se um triangulo identico a ADC sobre o lado BC de forma que BC=AD e que o angulo x esteja justaposto a C, temos:

Triangulo construido BCJ = triangulo ADC

assim o angulo CBJ= 135 graus.

Prolongue o segmento JB ate encontrar a reta AC no ponto P, agora note que o angulo CBP=ABJ= 45 graus. Com isso, os angulos APB e CPB sao iguais a 90 graus.

Tomando do se sem perda de generalidade que BJ= k temos: 2BJ=AC e AD=AB=BC= k . raiz quadrada de 2

Aplicando do se a Lei do Senos no triangulo ADC obtemos:

2k / seno de 135 graus = k . raiz quadrada de 2 / seno de 45-x

como x< 45

X= 15 graus.

Solution by: Joao Pedro Queroga

Bae deok rak (mexcalibur@naver.com)

ReplyDeleteSince AB=AD=BC, two triangles ABD, BCD are isoscele and angle (BAD)=45+x,

angle(ABD)={180-(45+x))/2 and so angle(CBD)=90-(180-(45+x))/2=(45+x)/2.

Therefore we see that 90-x=angle(BCD)=(180-(22.5+x/2))/2, that is, x=15.

Extend DB to E such that BE = BA = BC. Then B is the centre of the circumcircle of Tr. AEC. Hence < AEC = 45 ( 1/2 of < ABC = 90). But < ADC = 135 hence this circle passes thro' D too. This implies that Tr. ABD is equilateral from which x+45 = 60 and x = 15.

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

B is the circumcenter of triangle ACD, so BD=AB=AD, triangle ABD is equilateral, <ABD=45+x=60, x=15.

ReplyDeleteBest regards,

Stan Fulger

https://www.youtube.com/watch?v=WNj8tSyOK48

ReplyDeleteConstruct point E such that AE is perpendicular to EC, and triangle ACE is a right-angled triangle with <AEC=90.

ReplyDelete<ADC=135

Hence <ADE=45, and thus triangle AED is an isosceles right-angled triangle.

Let length of AE be 1.

Length of AD=SQRT(2) (Pyth. theorem)

AD=AB=BC

So AC=2 (Pyth. theorem)

Since AC=2 and AE=1, triangle AEC is a 30-60-90 triangle, with <ACE=30.

Hence, 45-x=30

x=15

See this video

ReplyDeleteDraw cercle circumscribed to ∆ABC, centered on H, midpoint of AC, since ∆ABC is iscoceles and rectangle at B.

ReplyDeleteExtend BH to cut circle at E, thus <ABE=45°

Extend CD to cut circle at F.

<ABF=<ACF=45°-x since subtending same arc AF, thus <FBE=x

Draw chord KP//AC. Then <AKP=<CAK=x since they are alternate-internal

Draw chord FL//PK. Then <FLP=<LPK=x since they are alternate-internal

Add the 3 angles: <ABE=45°=<AKP+<PLF+<FBE=3x x=15°