The bisector of an isosceles triangle, drawn from the vertex, is half the length of the other bisector.

Find the angles of the triangle.

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Let $A$ be the vertex of isosceles triangle $ABC$, and its bisector $AM$ is half the length of bisector $BD$. On the extension of bisector $AM$ beyond point $M$, we lay off segment $MK$ equal to $AM$. Then $BK \parallel AD$ and $AK = BD$. Therefore, $ADKB$ is an isosceles trapezoid. If $P$ is the point of intersection of the bisectors of triangle $ABC$, then $BP = KP$. Let $\angle ABP = \alpha$. Then $\angle PKB = \angle PBK = 3\alpha, \angle BAK = 90^\circ - 2\alpha, \angle BPK = \angle ABP + \angle BAK$, or $180^\circ - 6\alpha = \alpha + (90^\circ - 2\alpha)$.

From this, $\alpha = 18^\circ$. Therefore, $\angle B = 36^\circ$.

## Answer

$36^\circ, 36^\circ, 108^\circ$.