SOLUTION: ABCD is a parallelogram;
AB = 2BC, L − midpoint of DC , BL∩AD=M
Prove: ΔBCL ≅ ΔMDL, AL⊥BM, AL −∠ bisector of ∠A
This is pictu
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-> SOLUTION: ABCD is a parallelogram;
AB = 2BC, L − midpoint of DC , BL∩AD=M
Prove: ΔBCL ≅ ΔMDL, AL⊥BM, AL −∠ bisector of ∠A
This is pictu
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Question 1117901: ABCD is a parallelogram;
AB = 2BC, L − midpoint of DC , BL∩AD=M
Prove: ΔBCL ≅ ΔMDL, AL⊥BM, AL −∠ bisector of ∠A
This is picture:
http://homework.russianschool.com/resource?key=00nyq Answer by greenestamps(13203) (Show Source):
Let x be the length of BC, so the length of AB is 2x.
The length of AD is also x, because ABCD is a parallelogram.
The lengths of CL and DL are both x, becaue L is the midpoint of CD.
Angle BCL is congruent to angle MDL; angle CBL is congruent to angle DML (parallelogram; alternate interior angles).
Angles MLD and BLC are congruent (vertical angles).
Angle DML is congruent to angle CBL (angle sum of triangles DML and CBL).
The length of DM is x (congruent sides opposite congruent angles).
The length of AM is 2x.
Triangle MLA is congruent to triangle BLA (SSS).
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.
