See complete Problem 230 at:
gogeometry.com/problem/p230_triangle_midpoint_transversal_perpendicular.htm
Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Thursday, January 22, 2009
Elearn Geometry Problem 230: Triangle, Midpoint, Transversal
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One unimaginative but effective way of solving this problem is to let A be (0,0), B:(q,r) & C:(p,0) and t : y= mx + c or mx-y+c=0. By convention, distances above t are taken as positive and those below negative. We now have the following:D:(q/2,r/2), E:((p+q)/2,r/2)& F:(p/2.0). So d={mq/2-r/2+c)/(1+m^2)^(1/2)
ReplyDeletee=(m(p+q)/2 -r/2+c)/(1+m^2)^(1/2) while f =(-mp/2-c)/(1+m^2)^(1/2) since f is below the line t. Hence, d+e+f=(mq-r+c)/(1+m^2)^(1/2) which is nothing but the distance of B:(q,r) from t. In other words, b=(mq-r+c)/(1+m^2)^(1/2)
or b = d + e + f
Ajit: ajitathle@gmail.com
join D and F than DF=EC ( middle line ) ( 1 )
ReplyDeletedraw DF" // t and EB" // t
Tr DFF" and BB"E are similar ( DF // EC )
than
f+d / DF = b-e / EC from (1) f+d = b-e
or f+d+e = b
See the drawing
ReplyDelete- Define D’’ intersection of AC with DD’, B’’ with BB’, E’’ with EE’
A’ such as AA’ ⊥ D’E’ and C’ such as CC’ ⊥ D’E’
- Define d2=D’D’’, b2=B’B’’, e2=E’E’’, a=AA’, c=CC’
- D middle of AB=> D’’ middle of AB’’ => (1) 2d2=a+b2
- E middle of BC=> E’’ middle of B’’C => (2) 2e2=c+b2
- F middle of AC=> (3) 2f=c+a
- D middle of AB and E middle of BC => DE//AC
=>(4) d+d2=e+e2=(b+b2)/2
(5) d+e+f = ((b+b2)/2 –d2)+((b+b2)/2-e2)+(c+a)/2
(5) d+e+f =b+b2-d2-e2+(c+a)/2
- From (1) and (2) : (6) d2+e2=(a+c)/2+b2
(5) d+e+f =b+b2-(a+c)/2-b2+(a+c)/2
- Therefore d+e+f=b