## Tuesday, December 8, 2009

### Problem 399: Right triangle, Midpoints, Distance, Pythagorean theorem, Congruence

Proposed Problem
Click the figure below to see the complete problem 399 about Right triangle, Midpoints, Distance, Pythagorean theorem, Congruence.

Complete Problem 399
Level: High School, SAT Prep, College geometry

1. 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).
I 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

2. 1. Locate a point H such that CH=d and CH // AB
2. 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

3. http://img171.imageshack.us/img171/5231/problem399.png
See attached sketch of above problem
Peter Tran

4. Let the parallel thro' D to FG meet EG in H. Then DH = 2x by applying the mid point theorem to Tr. DEH.
Now 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