Saturday, October 10, 2009

Problem 363: Right triangle, Congruence, Angles

Proposed Problem
Click the figure below to see the complete problem 363 about Right triangle, Congruence, and Angles.

Problem 363. Right triangle, Congruence, Angles.
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Complete Geometry Problem 363
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 10:26 AM
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9 comments:

  1. AnonymousOctober 13, 2009 at 12:51 AM

    By considering isos triangle, alpha=45
    Then, by considering tan(beta+theta),
    tan(beta+theta)=(tan(beta)+tan(theta))/(1+tan(beta)tan(theta))
    =(1/2+1/3)/(1-1/2*1/3)
    =1
    So beta+theta=45,
    alpha+beta+theta=90

    ReplyDelete
  2. AnonymousNovember 2, 2009 at 2:37 PM

    How do you solve the problem without useing trigonometric approch???
    Just useing pure Geometry!!!!
    Is there anybody who has any idea???

    ReplyDelete
  3. bae deok rak(bdr@korea.com)October 2, 2010 at 5:43 PM

    Let D_1 be a point with ABDD_1 is a square and let D_2 be a point with DD_1 =DD_2 and BD is a perpendcular to D_1D_2.
    Then we have D_2A=D_2C and ang(D_2CD)=ang(AEB)=beta, that is, the triangle D_2AC is a right isoscele triangle. Therfore ang(D_2CA)=45=beta +gamma and alpha=45. We get
    ang(ADB)+ang(AEB)+ang(ACB)=alpha+beta+gamma=90 degrees.




    sum of the angles ADB, AEB, and ACB is 90 degrees.

    BDer

    ReplyDelete
  4. PrabhakarAugust 17, 2011 at 7:48 AM

    1.in tri ABD, ang.ADB=45 ...(iso. tri thm)

    2.hence alpha=45degree

    3.tri. DEA similar to tri CAD ...(sss test of simlarity)

    4.hence ang.AED=ang.DAC ...(c.a.s.t)

    5.hence ang.DAC=Beta ...(From 4)

    6.in triangle ADC,
    ang.DAC+ang.ACD=ang.ADB...(exterior ang. thm)

    7.hence beta+theta=45 degree...(from 5,6 & given)

    8.hence alpha+beta+theta=45+45=90 degree ...(from 2 & 7)

    Q.E.D

    ReplyDelete
  5. Sumith PeirisJune 28, 2015 at 10:56 AM

    Since alpha=45 we need to prove that beta + theta = 45.
    This can be seen by noting that AD^2=DE.DC. Hence AD is tangential to circle AEC and so < DAE = theta = 45 - beta. QED

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. Marvin KalnganOctober 12, 2017 at 6:18 AM

    Let AB=1.
    Since ABC is isosceles, alpha = 45 degrees.
    tan beta=1/2 and tan theta = 1/3
    so beta = arctan 1/2 and theta = arctan 1/3
    Therefore,
    45 + arctan 1/2 + arctan 1/3 = 90 degrees

    ReplyDelete
  7. GregJuly 27, 2020 at 11:49 AM

    Make ABDF a square and note that since ΔADE and ΔCDA are similar (SAS), <DAC = <DEA = β. Therefore <BAD = α, <DAC = β and <CAF = θ, hence α+β+θ =90°.

    ReplyDelete
  8. Sumith PeirisJuly 30, 2020 at 12:58 AM

    Another Geometry Proof

    Complete squares ABDX, XDEY, AXWZ

    In Tr.s ABE & AWC, BE/AB = CW/AW = 2. Further < ABE = AWC = 90

    It follows that the 2 Tr.s are similar & so < AEB = < ACW i.e Beta = 45 - Theta

    Hence Beta + Theta = 45 and since Alpha = 45 the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  9. rv.littlemanAugust 24, 2020 at 12:40 PM

    See the drawing: Drawing

    BA=BD and α = 45°
    Pythagoras : AD^2=2, AE^2=5, AC^2=10 => AC^2= 2AE^2
    This means that AC is the hypothenuse of a square AE x AE
    Inverse the drawing and translate it on the left of a distance BD
    Define G and F respectively the equivalents of B and A
    α = 45° => ΔABF is congruent to ΔDBF (SAS)
    => AF=DF
    DF=AE by construction => AF=AE
    Therefore ΔFAE is isosceles, and since AC is the hypothenuse of a square AE x AE => β+θ =45° => α+β+θ =90°

    ReplyDelete

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