Home > Shapes > Geometry
two2one (Posted on 2010-10-27) |
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Let AD be an altitude of triangle ABC.
Prove the following:
If /B = 2 /C < 90°, then |AB| + |BD| = |DC|.
Solution
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| Comment 1 of 2
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Let B’ be a point on CD such that |DB’| = |DB|.
/CAB’ = /AB’D - /C (Exterior angle of triangle) = /B - /C (Triangles ABD, AB’D congruent) = 2 /C - /C = /C
Therefore triangle AB’C is isosceles and |AB’| = |B’C|.
Hence |AB| + |BD| = |AB’| + |B’D| = |B’C| + |B’D| = |DC|
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Posted by Harry
on 2010-10-27 18:54:58 |
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