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°

  21. Grab a pen and a piece of paper, and illustrate along.

    Add point E into the figure and connect it to point B such line BE = line CD and angle CBE = 135°. Lastly connect point E to points A and C. You will see that two duplicates of triangle ACD are formed.

    But, how do we know they are identical?
    We can prove that they are identical if they are congruent, so we can use the SAS congruence test.

    Triangles BCE and ACD are congruent by SAS because:
    > Line AD = line BC
    > Angle ADC = angle CBE = 135°
    > Line CD = line BE

    Triangles ABE and BCE are congruent by SAS:
    > Line AB = line BC
    > Angle ABE = 360° - 90° - 135° = 135° = angle CBE
    > They share side BE.

    Since triangles ACD, ABE and BCE are congruent. lines AC, AE and CE are equal too. This means triangle ACE is equilateral. Since all three angles in an equilateral measure 60°, 45° + x = 60°.

    x = 60° - 45° = 15°

  22. Se construye la circunferencia circunscripta al triángulo ACD y es fácil notar que B es el circuncentro, como AB=AD entonces ABD es equilatero por lo que 45+a=60 a=15

  23. Draw perpendicular bisector of AC.Mark on it BE = DC. Connect p.A and p.E. Now triangle ABE is a duplicate of triangle ACD. Angle ABE = angle ADC = 135. But angle CBE is also equal to 135°.( 135°+135°+90°=360°) So tringle CBE is also duplicate of triamgoe ACD. Therefore triangle AEC is equiliteral. Angle ACB = 45° ( triangle ABC is isosceles). Angle BCE=x. Angle ACB=angle BCE=angle ACE = 60° -> 45°+x=60° and x=15°

  24. A very easy solution, no need to add any line or angle
    Let AB=BC=AD=a
    By Pyth. Theorem, AC=a*sqrt(2)
    By considering triangle ACD and using sine law