Sunday, May 18, 2008

Elearn Geometry Problem 8

Geometry Problem, Triangle, Angles

See complete Problem 8 at:

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

Post your solutions or ideas in the comments.


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

    Extend 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

  2. after bilding AE perpendicular with CD , a circle
    can 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

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


  5. bae deok rak( 3, 2010 at 5:32 PM

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

  6. Solution by Cristian

    Greetings :)

  7. solution is uploaded to the following link:


    Since ∠ (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

  9. In the usual notation c = a = AD and b² = 2a², b = a√2
    Denote 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°

  10. draw perpendicular to EC from A,E is point on extension of CE
    AC=a*sqrt2, angEDA=x+45x=45, angEAD=45, AD=a(given).so, AE=ED=a/sqrt2, AC=a*sqrt2, so,AC=2AE &, angACE=30=45-x

  11. To Sanoojan Baliah, problem 8.
    I 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?

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

  12. To Nilton Lapa

    Actually Sanoojan Baliah Says:
    Produce CD and Draw a perpendicular to CD from A it is AE.

    Greetings and THanks

  13. To Rengaraj PradheepKannah.
    Thanks again for your help. Now I understand the solution presented by Sanoojan Baliah, it is the same idea of Ahmetelmas.

  14. Solution- Solucao ( English- Portugues)
    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.

    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

  15. Bae deok rak( 20, 2015 at 4:25 PM

    Bae deok rak (
    Since 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.

  16. 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.

    Sumith Peiris
    Sri Lanka

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

    Best regards,
    Stan Fulger


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

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