M22-02

Official Solution:


If the diagonals of rectangle \(ABCD\) intersect at point \(E\), what is the degree measure of \(\angle AEB\) ?



(1) The degree measure of \(\angle EDC\) is 15º.

When two parallel lines are cut by a transversal, alternate interior angles are equal. Therefore, angles EDC and ABE are equal since AB and DC are parallel and intersected by DB. Thus, we know that angle EDC is 15 degrees.

Next, since ABCD is a rectangle, its diagonals bisect each other, meaning that AE = BE. Thus, angle ABE = angle BAE = 15 degrees.

Finally, in triangle ABE, we know that angles ABE, BAE, and AEB add up to 180 degrees. Since ABE and BAE are both 15 degrees, we can solve for AEB to be 150 degrees.

Therefore, the first statement is sufficient.

(2) The degree measure of \(\angle ACB\) is 75º.

Since ABC is a right triangle, we know that angle ACB and angle BAC add up to 90 degrees. Therefore, angle BAC is 15 degrees. Finally, in triangle ABE, we know that angles ABE, BAE, and AEB add up to 180 degrees. Since ABE and BAE are both 15 degrees, we can solve for AEB to be 150 degrees.

Therefore, the second statement is sufficient.


Answer: D
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The problem says it's a rectangle. A rectangle's diagonals bisect each other and are equal in length, so if you draw a rectangle and draw in the diagonals, you'll get 4 isosceles triangles. Both statements give one angle of the two angles between the base and the equal sides, so you can calculate the third angle from both statements and get to AEB.

Bunuel. Please post a pictorial explanation. Thanks.

Here it is:



Does it help?



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No Bunuel I could not relate the picture with your solution given earlier. Could you please explain it with picture how \(\angle AEB = 180 - 2*\angle EDC = 150^o\)?

and \(\angle AEB = 180 - 2*\angle BAC = 180 - 2*(180 - 90 - \angle ACB) = 150^o\)?

For (1). First of all, \(\angle AEB = \angle DEC\). Next, DEC is isosceles, thus \(\angle EDC = \angle ECD=15^o\). The sum of the angles in a triangle is 180 degrees:

\(\angle EDC + \angle ECD + \angle DEC = 180^o\);

\(\angle EDC + \angle EDC + \angle AEB = 180^o\);

\(\angle AEB = 180 - 2*\angle EDC = 150^o\).


For (2). First of all, \(\angle BAC + \angle ACB= 90^o\) --> \(\angle BAC =90^o-\angle ACB\). Next, AEB is isosceles, thus \(\angle BAC = \angle ABE\). The sum of the angles in a triangle is 180 degrees. \(\angle BAC + \angle ABE + \angle AEB = 180^o\);

\(\angle BAC + \angle BAC + \angle AEB = 180^o\);

\(\angle AEB =180^o-2*\angle BAC = 180^o-2*(90^o-\angle ACB)\)
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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