Geometry Problem

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

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## Thursday, November 25, 2010

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

Labels:
90,
altitude,
angle bisector,
degree,
perpendicular,
right triangle

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▲BFG isoceles from 541 => BG altitude

ReplyDeleteYou mean triangle BHE is isosceles (where BE, AD meet at H)?

ReplyDelete▲BEH isoceles, AE meet BD at H

ReplyDelete/_ABD = 90-/_A or/_DBC=/_A so/_DBG = /_A/2

ReplyDeleteThus,/_DAG=/_A/2=/_DBG or A,B,G & D are concyclic; hence x = /_ADB =90.

Ajit

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

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

Another method

ReplyDelete< ABG = < AFG and the result follows