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Incircle Fixed Point (Posted on 2010-12-08) |
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Let A, B, and P be three fixed, non-collinear points.
Let C be a point on ray AP different from A.
The incircle of ΔABC touches BC at D and AC at E.
Prove that the line DE passes through a fixed point as C varies on ray AP.
Solution
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Comment 1 of 1
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Let |BD| = b and |CD| = |CE| = c. Let F be a fixed point on AC such that |AF| = |AB|, then |EF| = b. Let M be the mid-point of BF. Let d and e be unit vectors in the directions of CD and CE.
Vector EM = 0.5 (EF + EB) = 0.5 (EF + EC + CD + DB) = 0.5 [(|EF| + |EC|)(-e) + (|CD| + |DB|)d] = 0.5 [(b + c)(-e) + (c + b)d] = 0.5 (b + c)(d - e) = 0.5 (b/c + 1)(cd - ce) = 0.5 (b/c + 1)(CD - CE) = 0.5 (b/c + 1) ED
Thus EM is parallel to ED, and the fixed point M must always lie on ED, whatever the values of b and c; i.e. for all positions of C on the ray AP.
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Posted by Harry
on 2010-12-11 00:53:18 |
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