Friday, February 7, 2014

Geometry Problem 975: Isosceles Right Triangle, 45 Degrees, Angle Bisector, Hypotenuse

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

Click the figure below to see the complete problem 975.

Online Geometry Problem 975: Isosceles Right Triangle, 45 Degrees, Angle Bisector, Hypotenuse

4 comments:

  1. Let E be a point on AC such that DE⊥AC.

    Then ΔABD congruent to ΔAED (AAS).
    So AE=AB.

    Now ΔEDC is an isosceles right triangle.
    Thus EC=ED=BD.

    Hence, AC=AE+EC=AB+BD.

    ReplyDelete
  2. Point E is on AB extended such that EB=BD. Then <BED=<DCA=45, so triangles ADE and ADC are congruent. Therefore AC=AE=AB+BE=AB+BD

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  3. AC=V2 . AB ; CD=V2. BD (angle bisect law)
    BC=BD+CD= BD+V2.BD => BC+BD=2BD+V2.BD
    BC=BD+V2.BD =>AC=V2.BC=V2.BD+2BD => AC=BC+BD

    ReplyDelete
  4. Extend AB to E such that AE = AC

    Then Tr.s ABD and BCE are congruent ASA and so BD = BE

    Hence AC = AE = AB + BE = AB + BD

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete