In a triangle ABC, D is a point on BC such that AD is the internal bisector of angle A. Now Suppose angle B = 2 * angle C and CD=AB. Prove that angle A=72°.
Let CD=AB=1. Call AC=x. By the angle bisector theorem, DB=1/x.
Bisect angle ABC and call the intersection with AD point E. Let AE=y.
Triangles ACD and ABE are similar (each having angles of half of A and half of B) from which EB=1/x.
Using the angle bisector theorem again, ED=y/x, so AD=y+y/x.
Using the similar triangles again, y+y/x=xy or 1+1/x=x.
So BC=AC.
Triangle ACB is isosceles with vertex angle equal to half of the base angles. Angle A=72.
QED
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Posted by Jer
on 2025-05-17 18:26:48 |
Proof