ABCD is a parallelogram. L and M are points on AB and DC respectively and AL=CM. Prove that LM and BD bisect each other.
Let the intersection of BD and LM be P.
Now,
AL = CM so BL = DM
m<ABD = m<CDB
m<BLM = m<DML
So, by ASA, LBP and MDP are congruent triangles.
S, LP = PM and DB bisects LM
Similarly, DP=BP and LM bisects BD
Vuvyuyv
To prove that LM and BD bisect each other in parallelogram ABCD, we can use the concept of triangles and their properties. Here's a step-by-step proof:
Step 1: Given that ABCD is a parallelogram.
We know that opposite sides of a parallelogram are parallel and congruent, which means AB || DC and AB = DC.
Step 2: Given that AL = CM.
It is given that AL = CM, which means the segments AL and CM have equal lengths.
Step 3: Draw diagonal AC.
Draw the diagonal AC to divide the parallelogram into two triangles, ΔABC and ΔADC.
Step 4: Prove ΔABC ≅ ΔADC.
Since AB || DC, and AC is a transversal, we have ∠A = ∠D (corresponding angles).
Also, AB = DC (given), and AL = CM (given).
Therefore, by the Side-Angle-Side (SAS) congruence condition, we have ΔABC ≅ ΔADC.
Step 5: Prove ΔBLC ≅ ΔCMD.
Since ΔABC ≅ ΔADC, we have ∠C = ∠B (corresponding angles) and BC = CD (corresponding sides).
Since AL = CM (given), we have LB = DM (subtracting equal lengths).
Therefore, by the Side-Angle-Side (SAS) congruence condition, we have ΔBLC ≅ ΔCMD.
Step 6: Prove LM || BD.
We know that ΔBLC ≅ ΔCMD, so ∠BLM = ∠CMD (corresponding angles) and LM = MD (corresponding sides).
Since the opposite sides of parallelogram ABCD are parallel, we have ∠DAB = ∠ABC = ∠LBM (alternate interior angles).
Therefore, LM || BD.
Step 7: Prove LM and BD bisect each other.
Since LM || BD, and ∠BLM = ∠CMD (proven in Step 6), LM and DB intersect at point O.
Using the Alternate Interior Angles (AIA) theorem, we can conclude that LM and BD bisect each other at point O.
Thus, we have proven that LM and BD bisect each other in parallelogram ABCD.
To prove that LM and BD bisect each other, we need to show that their point of intersection is the midpoint of both segments.
Let's start by drawing a diagram to visualize the given information.
```
A_______B
|\ /|
| \ / |
L___|__\/__|___M
| /\ |
| / \ |
|/____\|
C D
```
We know that ABCD is a parallelogram, so the opposite sides are parallel and equal in length.
Now, consider triangles ALC and CMD. We are given that AL = CM, and they share an angle at L and M, respectively. Moreover, both triangles have one side in common, AC and CD, respectively.
By the Side-Angle-Side (SAS) congruence criterion, we can conclude that triangles ALC and CMD are congruent. Therefore, ∠ALC = ∠CMD and ∠ALM = ∠CMD.
Since the opposite angles are equal, we deduce that AL || MD and LM || AD.
Now, we have two pairs of parallel sides: AD || LM and AL || MD.
According to the Converse of the Alternate Interior Angles Theorem, if two lines are parallel and cut by a transversal, then the pairs of alternate interior angles are congruent. Thus, we can conclude that ∠ALM = ∠MDB.
Since ∠ALM = ∠MDB, we can infer from vertical angles that ∠AMC = ∠LDB.
Considering we have two pairs of congruent angles (AL = CM and ∠AMC = ∠LDB) and one pair of opposite sides parallel (LM || AD), we can apply the Angle-Angle (AA) similarity criterion for triangles AMC and LDB.
As a result, we can state that triangles AMC and LDB are similar.
In similar triangles, the corresponding sides are in proportion. Therefore, LM/BD = AM/AD.
Since ∠AMC = ∠LDB, we can conclude that LM and BD bisect each other at their point of intersection. Thus, the point of intersection is the midpoint of both segments.
Hence, we have proven that LM and BD bisect each other.