OC693 from Crux Mathematicorum Vol. 50, No. 7

OC693 (from Crux Mathematicorum Vol. 50, No. 7)

On the side \(AC\) of triangle \(ABC\), point \(E\) is chosen. Bisector \(AL\) intersects segment \(BE\) at point \(X\). It turns out that \(AX = XE\) and \(AL = BX\). What is the ratio of angles \(A\) and \(B\) of the triangle?

Solution

The ratio of angles \(A\) and \(B\) of the triangle is \(2:1\).

Proof

From angle bisector theorem, we have \(AB:BX = AE:XE\). Since \(AX = XE\) and \(AL = BX\), we get \(AB:AL = AE:AX\). Since \(\angle BAL = \angle XAE\), we have \(\triangle LAB \sim \triangle XAE\), which implies \(LA = LB\), \(\angle ABL = \angle BAL\). Therefore \[\angle A = 2\angle BAL = 2\angle ABL = 2\angle B.\]