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

Labels:
congruence,
distance,
midpoint,
Pythagoras,
right triangle

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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