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Triangles AED and BEC are formed using the straight lines AB and CD as shown in the figure above. If \(BE = BC, DE^2 > AD^2 + AE^2\) and the measure of \(∠AED\) is x, which of the following statements must be true?
I. \(CE^2 > 2BE^2\) II. \(AE < AD\) III. \(DE > CE\)
A) I only
B) II only
C) III only
D) I, II and III
E) None of the above
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Let's label the figure from the given information:
1.1) \(∠DAE\) = (180-3x) from lower triangle
1.2) Since E is the intersection point of two lines, \(∠BEC\) = \(∠AED\) = x
1.3) Since BE = BC, \(∠ECB\) = \(∠BCE\) = x
1.4) \(∠EBC\) = 180-2x
1.5) Since \(DE^2 > AD^2 + AE^2\), we know that angle opposite to side DE is > 90 degrees and it is an obtuse triangle. So angle opposite to DE is \(∠DAE\) which from 1.1 is 180-3x.
So 180-3x >90, and x <30
2) Evaluating each statement independently:
Now that we have established our pre-thinking, let's evaluate each statement:
I) \(CE^2 > 2BE^2\) which is the same as \(CE^2 > BE^2 + BC^2\) since BE = BC
Now for the above inequality to be true, angle opposing CE should be greater than 90, which means 190-2x should be greater than 90 , we already know that x <30, so 180-2x should be less than 120.
Hence sufficient condition. I is true.
II) AE <AD
we know that angle opposing AD is greater than angle opposing side AE, so the above inequality cannot be true.
III) AE > AD
we know that angles opposing each side is proportional to that side length. So angle opposing AE = 180-3x and angle opposing AD is 180-2x. Under all circumstances. 180-3x < 180-2x,
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.