Tuesday, October 27, 2009

Problem 377: Triangle, Internal Angle Bisector, Altitude

Proposed Problem
Click the figure below to see the complete problem 376 about Triangle, Internal Angle Bisector, Altitude.

 Problem 377: Triangle, Internal Angle Bisector, Altitude.
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Complete Geometry Problem 377
Level: High School, SAT Prep

2 comments:

  1. A+B+C=180 and B/2+alpha=90-A.
    So, (180-A-C)/2+alpha=90-A,
    alpha=(C-A)/2.

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  2. alpha = B/2-(90-C)=B/2-A/2-B/2-C/2+C=(C-A)/2

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