perplexus.info :: Geometry : More Collinear Points

More Collinear Points (Posted on 2009-12-11)

Let PQ be a diameter of a circle,
     A be a point on line PQ such that P lies between A and Q,
     T be a point on the circle such that line AT is tangent to the circle,
     B be the point on line QT such that line BP is perpendicular to line PQ, and
     C be the point on line PT such that line CQ is perpendicular to line PQ.

Prove that points A, B, and C are collinear.

Submitted by Bractals

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Solution: (Hide)

Let O be the center of the circle and point S the perpendicular projection of point T on line PQ.

From the similar triangles QST and QPB,
|ST| |PB| ------ = ------ (1) |QS| |QP|
From the similar triangles PST and PQC,
|ST| |QC| ------ = ------ (2) |SP| |QP|
Combining (1) & (2) we get,
|SP| |PB| ------ = ------ (3) |QS| |QC|
From the similar triangles OTA and OST,
|OT| |OS| ------ = ------ (4) |OA| |OT| (|OT|+|OA|)*(|OT|-|OS|) = (|OT|+|OS|)*(|OA|-|OT|) (|QO|+|OA|)*(|OP|-|OS|) = (|QO|+|OS|)*(|OA|-|OP|) |QA|*|SP| = |QS|*|PA| |SP| |PA| ------ = ------ (5) |QS| |QA|
Combining (3) & (5) we get,
|PB| |QC| ------ = ------ (6) |PA| |QA|
Therefore, triangles APB and AQC are similar.

Thus, points A, B, and C are collinear.

See Harry's post for an alternate proof.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Solution	Harry	2009-12-13 11:59:59
re(2): counterproof (maybe)	Daniel	2009-12-12 16:00:17
re: counterproof (maybe)	Charlie	2009-12-12 12:16:10
counterproof (maybe)	Daniel	2009-12-12 10:37:21

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