Let ABC be a triangle with P a point on line BC
such that B lies between P and C.
Construct a line through P, intersecting sides
AB and AC in points Q and R respectively, such
that vertex S of parallelogram QARS lies on
line BC.
(In reply to
Solution by Brian Smith)
Bravo Brian. I liked your algebraic interlude in paragraph 3, which revealed the way of constructing S. When I saw this need for PS/PC = PB/PS it prompted me to think about the tangent-chord property of a circle which led to the following slightly different construction.
Draw the circle with BC as diameter and call its centre M. Then draw the circle with PM as diameter crossing the first circle at D. Now draw a circle centred on P and passing through D. This circle crosses BC at S.
Proof: Angle PDM is a right angle (subtended by diameter PM) so PD is a tangent to the circle BCD and, in conjunction with the secant PC, gives:
PS2 = PD2 = PB.PC, showing that S is the required point.
Edited on July 13, 2010, 8:02 pm
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Posted by Harry
on 2010-07-13 20:00:54 |
re: Solution
