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In triangle ABC, side AB is extended to point D. If the bisectors of angle CAB and angle CBD meet at point E, prove that angle AEB is equal to half of angle ACB.

A. Angle AEB = 1/2 * Angle ACB
B. Angle AEB = Angle ACB
C. Angle AEB = 2 * Angle ACB
D. Angle AEB = 3 * Angle ACB

Answer :

Final answer:

In triangle ABC, if the bisectors of angle CAB and angle CBD meet at point E, angle AEB is equal to half of angle ACB. This is because the bisectors divide the angles into equal parts, and the sum of angles in a triangle equals 180 degrees.

Explanation:

In triangle ABC, given that the bisectors of angle CAB and angle CBD meet at point E, let's try to find the value of angle AEB.

Since CE is an angle bisector, it divides angle ACB into two equal parts. Therefore, angles ACE and BCE are rqual. Let's denote the measure of these angles as x.

Similarly, BE is the angle bisector of angle CBD, so it splits this angle into two equal parts as well.

Let's denote them as y.

Hence, Angle ABD is equal to angle CAB + angle CBD which is equal to x + 2y.

Since the sum of angles in any triangle is 180 degrees, we find that in triangle AEB, angle AEB = 180 - (x+y) = 180 - 1/2 angle ABD. Therefore, angle AEB is indeed equal to half of angle ACB (because angle ACB = 2* angle ABD).

Learn more about Angle Bisectors in Triangles here:

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