MHB -gre.ge.2 distance by similiar triangles

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The discussion focuses on using similar triangles to find the length x in a geometry problem involving a summer camp counselor. Given the lengths AB, EB, BD, and CD, the relationship between these segments is established through the angles AEB and CDE being equal. The calculations show that x can be determined using the proportion x/EB = CD/BD, leading to the conclusion that x equals 1600 ft. A correction is noted regarding the value of BD, confirming it should be 700 ft instead of 7000 ft. The final consensus is that x is indeed 1600 ft based on the corrected measurements.
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A summer camp counselor wants to find a length, x,
The lengths represented by AB, EB BD,CD on the sketch were determined to be 1800ft, 1400ft, 7000ft, 800 ft respectfully
Segments $AC$ and $DE$ intersect at $B$, and $\angle AEB$ and $\angle CDE$ have the same measure What is the value of $x$?

looks easy but still tricky

$\dfrac{x}{EB}=\dfrac{CD}{BD}
=\dfrac{x}{1400}=\dfrac{800}{700}$
multiple thru by 1400 then simplify
$x=\dfrac{800(1400)}{700}=(800)(2)=1600$ hopefully

I thot I posted this problem some time ago here but didn't see the solution in my overall document🕶
 
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x = 1600 ft only if BD = 700 ft instead of 7000 ft
 
mahalo good catch
yes I think its 700 not 7000
 

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