Proposed Problem
Click the figure below to see the complete problem 399 about Right triangle, Midpoints, Distance, Pythagorean theorem, Congruence.
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Complete Problem 399
Level: High School, SAT Prep, College geometry
Tuesday, December 8, 2009
Problem 399: Right triangle, Midpoints, Distance, Pythagorean theorem, Congruence
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This is a repeat of an earlier problem. Nonetheless, complete the rectangle ABCH and join D to F. Extend DF to meet CH in I. It's evident that DF=FI and hence triangles ADF & CIF are congruent which means that AD=CI=d. Since G & I are midpoints of DE & DI resply., we've 2GF=EI or 2x = √((CI^2+CE^2) = √(d^2+e^2).
ReplyDeleteI do not recall the earlier problem number but remember that Newzad had posted a similar or the same solution in his blog.
Ajit: ajitathle@gmail.com
1. Locate a point H such that CH=d and CH // AB
ReplyDelete2. Note that EH= SQRT (e^2 +d^2)
3. Note that Triangle ADG congruent with triangle CHG (case SAS) .
4. From step 3 we have Angle AGD= angle CHG and 3 points D, G, H are collinear
5. From triangle EDH, FG= ½ EH= ½ SQRT(e^2+d^2)
Peter Tran
Email: vstran@yahoo.com
http://img171.imageshack.us/img171/5231/problem399.png
ReplyDeleteSee attached sketch of above problem
Peter Tran
Let the parallel thro' D to FG meet EG in H. Then DH = 2x by applying the mid point theorem to Tr. DEH.
ReplyDeleteNow the diagonals of AECH bisect each other hence the same is a parellelogram and so AH = e and < BAH = 90.
So by applying Pythagoras to Tr. ADH the result follows
Sumith Peiris
Moratuwa
Sri Lanka