<PRE><B>Question 1117901</B>
&lt;br&gt;
Let x be the length of BC, so the length of AB is 2x.&lt;br&gt;
The length of AD is also x, because ABCD is a parallelogram.&lt;br&gt;
The lengths of CL and DL are both x, becaue L is the midpoint of CD.&lt;br&gt;
Angle BCL is congruent to angle MDL; angle CBL is congruent to angle DML (parallelogram; alternate interior angles).&lt;br&gt;
Angles MLD and BLC are congruent (vertical angles).&lt;br&gt;
Angle DML is congruent to angle CBL (angle sum of triangles DML and CBL).&lt;br&gt;
The length of DM is x (congruent sides opposite congruent angles).&lt;br&gt;
The length of AM is 2x.&lt;br&gt;
Triangle MLA is congruent to triangle BLA (SSS).&lt;br&gt;
The congruency of those two triangles tells us that AL is perpendicular to BM and that AL is the bisector  of angle A.  I leave the details to you for that last part of the proof.</PRE>