Question 62354: Right angle triangle that has a line drawn parallel to the base line of the triangle.
The top angle of the triangle is A, the right angle is B, and the last angle is C. There is a line about 2/3 of the way down between A and B, creating D.
This line segment will be called DE it runs parallel to BC.
AD=4, DB=2, BC=8. I figured out that AC=10 using the pythagorean theorem. But I can't figure out how to get DE or AE or EC. I've tried all sorts of equations and can't come up with an answer.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Right angle triangle that has a line drawn parallel to the base line of the triangle.
The top angle of the triangle is A, the right angle is B, and the last angle is C. There is a line about 2/3 of the way down between A and B, creating D.
This line segment will be called DE it runs parallel to BC.
AD=4, DB=2, BC=8. I figured out that AC=10 using the pythagorean theorem. But I can't figure out how to get DE or AE or EC. I've tried all sorts of equations and can't come up with an answer.
YOU ARE CORRECT USING PYTHOGARUS THEOREM
AC^2=AB^2+BC^2=6^2+8^2=100
AC=10
TO FIND DE,AE,EC, WE USE PROPERTIES OF SIMILAR TRIANGLES.IF YOU ARE NOT CONVERSANT WITH THAT THEN USE THE FOLLOWING METHOD
IN TRIANGLE ABC
DE||BC.....GIVEN.HENCE
AD/DB=AE/EC
AE/EC=4/2=2
AE=2EC
AC=10=AE+EC=2EC+EC=3EC
EC=10/3
AE=AC-EC=10 - 10/3 = 20/3
SINCE DE||BC,ANGLE ADE = ANGLE ABC=90....CORRESPONDING ANGLES.
SO IN RIGHT ANGLED TRIANGLE ADE,
AD=4,AE=20/3..HENCE
AD^2+DE^2=AE^2
16+DE^2=400/9
DE^2=400/9 -16 = 256/9
DE= 16/3
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