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| Parallel Midpoints (Posted on 2016-07-30) |
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Let BD be an altitude of ΔABC with D between A and C (not the midpoint).
The bisector of ∠ACB intersects BD and AB in points P
and Q respectively. The bisector of ∠CAB intersects BD
and BC in points R and S respectively. F and G are the
midpoints of PQ and RS respectively.
Prove that FG is parallel with AC if and only if ∠ABC = 90°.
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Submitted by Bractals
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Solution:
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To solve the problem I needed four additional points:
F', G', Q', and S'. These are the feet of the perpendiculars
from points F, G, Q, and S to the side AC.
The following are lowercase letters I assigned to the
line segments of the problem:
a = |BC| h = |BD|
b = |AC| p = |PD|
c = |AB| q = |QQ'|
d = |AD| r = |RD|
e = |CD| s = |SS'|
How an angle bisector cuts the opposite side of a triangle:
p = e*h/(a + e)
(1)
The point Q lies on side AB and angle bisector CP:
q/(b - |CQ'|) = |QQ'|/|AQ'| = |BD|/|AD| = h/d
q/(|CQ'|) = |QQ'|/|CQ'| = |PD|/|CD| = p/e
Solving these two for q,
q = b*h*p/(d*p + e*h)
(2)
Combining (1) & (2):
q = b*h/(a + b)
Therefore,
p + q = h*[b/(a + b) + e/(a + e)]
(3)
From right triangles ADB and CDB,
a2 = e2 + h2
c2 = d2 + h2 = (b - e)2 + h2
e = (b2 + a2 - c2)/(2*b)
(4)
Combining (3) & (4):
p + q = h*[b/(a + b) + (a2 + b2 - c2)/([a + b]2 - c2)]
(5)
Swapping the letters a, p, and q in (5) with the letters
c, r, and s respectively gives
r + s = h*[b/(c + b) + (c2 + b2 - a2)/([c + b]2 - a2)]
(6)
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FG || AC ⇔ |FF'| = |GG'|
⇔
p + q = r + s
⇔
p + q - r - s = 0
⇔
h*[b/(a + b) + (a2 + b2 - c2)/([a + b]2 - c2)] -
h*[b/(c + b) + (c2 + b2 - a2)/([c + b]2 - a2)] = 0
⇔
b*h*(a - c)(a + 3*b + c)(b2 - a2 - c2)
------------------------------------------------------- = 0 (7)
(a + b)(b + c)(a + b - c)(a + b + c)(b + c - a)
⇔
b2 = a2 + c2
⇔
∠ABC = 90°.
=============================================
QED
Note 1: The Factor function of Mathematica was used to get (7).
Note 2: The factor (a - c) in (7) was the reason that D was
restricted from being the midpoint of AC.
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