Saturday, February 8, 2014

Geometry Problem 976: Isosceles Right Triangle, 45 Degrees, Incenter, Angle Bisector, Hypotenuse

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 976.

Online Geometry Problem 976: Isosceles Right Triangle, 45 Degrees, Incenter, Angle Bisector, Hypotenuse

Posted by Antonio Gutierrez at 5:39 AM
Labels: , , , , ,

9 comments:

  1. Jacob HA (EMK2000)February 8, 2014 at 7:00 AM

    Let D be a point on BC such that AD bisects ∠A.
    Then AID is straight line.

    ∠BID=∠BAI+∠ABI=67.5°
    ∠BDI=90°−∠BAD=67.5°
    Thus, BI=BD.

    Hence, by Problem 975,
    AC=AB+BD=AB+BI.

    ReplyDelete
  2. Peter TranFebruary 8, 2014 at 8:09 AM

    http://imageshack.com/a/img716/8509/6hr3.png

    On BC locate E such that BI=BE
    Triangle IBE is isosceles with ∠ (IED)=22.5 degrees= ∠ (ICD)
    Triangles IDC and IDE are congruent…. ( Case SAS) => DC=DE
    And CA= 2 x CD=CE => CA=CB+BE= AB+BI

    ReplyDelete
  3. Sumith PeirisSeptember 16, 2015 at 11:05 PM

    The angle bisector BI divides isoceles right Tr. ABC. into 2 congruent isoceles right Tr. s.

    So if the in radius be r,
    AB - r = AC / 2 and BI + r = AC/2. Adding these 2 equations the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. APOSTOLIS MANOLOUDISDecember 7, 2016 at 2:43 AM

    Problem 976
    Let point D in the extension of AB to B such that AD=AC.Then triangle ADI and ACI are congruent (AD=AC,AI=AI,<DAI=<CAI=22.5 ).So <BDI=<AC[=22.5 and <ABI=<BDI+<BID or
    45=22.5+<BID or <BID=22.5 therefore BI=BD and AC=AD=AB+BD=AC+BI.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE



    ReplyDelete
  5. AnonymousDecember 7, 2016 at 7:36 AM

    (More of an algebric solution)
    Let AB=BC=1 => AC = √2 and BI extended meet AC at P
    Since m(BPC) = 90, triangle BPC is isosceles right angled
    => BP = CP = 1/√2
    Semi-perimeter of ABC , S=(2+√2)/2
    S.r=.5*AB.BC => r=IP=1/(2+√2)
    => BI = BP-IP = 1/√2 - 1/(2+√2)
    => BI = 2/2+2√2
    => BI = 1/1+√2 = √2-1
    => BI = AC-AB
    => AC = AB+BI

    ReplyDelete
  6. testDecember 7, 2016 at 9:07 PM

    2r=a+b-c
    sqrt(2)BI=AB + BC - Ac =2AB-sqrt(2)AB
    so sqrt(2) (BI+AB)=2AB=sqrt(2)AC
    so BI+AB=AC

    ReplyDelete
  7. Sumith PeirisJanuary 28, 2021 at 9:16 AM

    Solution 2

    Draw circle BIC to meet AB extended at X

    Then < ICX = 45 so < BCX = 22.5 = < ICB

    So BI = BX and AX = AC and the result follows:-

    Sumith Peiris
    MOratuwa
    Sri Lanka

    ReplyDelete
  8. Sailendra TJanuary 28, 2021 at 2:29 PM

    Let D,E be the points of tangency on AC and AB
    AD=DC=DB=Radius of circumcircle of ABC
    AC=2BD=2BI+2ID=2BI+2EB=2BI+2AB-2AE
    AC=2BI+2AB-2AD
    2AC=2BI+2AB
    AC=BI+AB

    ReplyDelete
  9. rv.littleman@heptic.frJune 25, 2021 at 5:04 AM

    See the drawing

    Define r as the inradius of ΔABC
    in a right triangle, r=(AB+BC-AC)/2
    AB=AC => r=(2AB-AC)/2=AB-AC/2
    BI=r.sqrt(2)=sqrt(2)(AB-AC/2)=sqrt(2).AB-AC/sqrt(2)
    AC=sqrt(2).AB => BI=AC-AB
    Therefore AC=AB+BI

    ReplyDelete

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