Wednesday, November 11, 2020

Geometry Problem 1483: Isosceles Right Triangle, Excenter, Perpendicular, Measurement

Geometry Problem 1483. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1483: Isosceles Right Triangle, Excenter, Perpendicular, Measurement.

Posted by Antonio Gutierrez at 4:21 AM

5 comments:

  1. Peter TranNovember 11, 2020 at 3:54 PM

    https://photos.app.goo.gl/h5V695JxQirG1RF98

    Define point F per sketch
    We have ∠ (ADF)= ∠ (ADB)= pi/8
    FD= 15*cos(pi/8)
    BD=FD*sqrt(2)
    EB=BD*sin(pi/8)
    So EB=15/2*sqrt(2)*sin(pi/4)
    EB=15/2

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    Replies
    1. Peter TranNovember 13, 2020 at 8:55 PM

      https://photos.app.goo.gl/mVTbQNZAG1vg6GvU9
      Geometry solution: Draw line DFP perpendicular to BA ( see sketch)
      Note that triangle BDP is isosceles and BP= 2.BE
      Triangle BDF is isosceles right triangle
      Triangle BFP congruent to DFA ( case ASA) so
      AD=BP=15 so BE= ½ BP= 7.5

      Delete
  2. c.t.e.o.November 12, 2020 at 12:49 AM

    Join D to B, D to C. Draw DH perpend to BC, DK perpend to ED.
    => Tr DBK isosceles. Draw BM perpend to DK => BM median
    => DM = EB = 7.5

    ReplyDelete
  3. Sumith PeirisDecember 15, 2020 at 7:33 AM

    Complete square BXDY. Let BD, AC cut at Z.

    Tr.s ADX, ADZ, ZDC & CDY are all congruent and Tr. ADX is also congruent with BXZ, ASA

    So BE = BZ/2 = AD/2 = 7.5

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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