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

Labels:
distance,
midpoint,
perpendicular,
transversal,
trapezoid,
triangle

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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