Thursday, November 25, 2010

Problem 542: Right Triangle, Altitude, Angle Bisector, Perpendicular, 90 Degrees

Geometry Problem
Click the figure below to see the complete problem 542 about Right Triangle, Altitude, Angle Bisector, Perpendicular, 90 Degrees.

Problem 542: Right Triangle, Altitude, Angle Bisector, Perpendicular, 90 Degrees.
Go to Complete Problem 542

6 comments:

  1. ▲BFG isoceles from 541 => BG altitude

    ReplyDelete
  2. You mean triangle BHE is isosceles (where BE, AD meet at H)?

    ReplyDelete
  3. ▲BEH isoceles, AE meet BD at H

    ReplyDelete
  4. /_ABD = 90-/_A or/_DBC=/_A so/_DBG = /_A/2
    Thus,/_DAG=/_A/2=/_DBG or A,B,G & D are concyclic; hence x = /_ADB =90.
    Ajit

    ReplyDelete
  5. < DBA = A hence the bisected angles are also equal. Hence ABGD is cyclic and so x = < BDA = 90.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. Another method

    < ABG = < AFG and the result follows

    ReplyDelete