Consider EDM and BCM. Since AD BC and AD is extended to E, AE BC, which gives alternate interior angles EDM = BCM. Point M is the midpoint of CD, so DM = CM, and vertically opposite angles are equal: EMD = BMC. By the ASA congruence criterion, EDM BCM, which yields ED = BC.

Consider EDM and BCM. Since AD BC and AD is extended to E, AE BC, which gives alternate interior angles EDM = BCM. Point M is the midpoint of CD, so DM = CM, and vertically opposite angles are equal: EMD = BMC. By the ASA congruence criterion, EDM BCM, which yields ED = BC.

Now consider AEL and CBL. Since AE BC, alternate interior angles give EAL = BCL and AEL = CBL. By the AA similarity criterion, AEL CBL. Therefore, the corresponding sides are proportional: (EL)/(BL) = (AE)/(CB).

Now consider AEL and CBL. Since AE BC, alternate interior angles give EAL = BCL and AEL = CBL. By the AA similarity criterion, AEL CBL. Therefore, the corresponding sides are proportional: (EL)/(BL) = (AE)/(CB).

Substitute AE = 2BC into the ratio from the similarity of triangles. This gives (EL)/(BL) = (2BC)/(BC) = 2. Multiplying both sides by BL yields the desired result: EL = 2BL.

$ABCD$ is a parallelogram. A line segment $BE$ bisects side $CD$ at $M$ and intersects diagonal $AC$ at $L$ , with point $E$ lying on the extension of side $AD$ . Prove that $EL = 2 BL$ . Figure Details: - Parallelogram $ABCD$ with vertices $A, B, C, D$ . - Side $AD$ is extended to a point $E$ . - Line segment $BE$ is drawn, intersecting side $CD$ at point $M$ such that $DM = MC$ ( $M$ is the midpoint of $CD$ ). - Diagonal $AC$ intersects the line segment $BE$ at point $L$ .

Answer: Hence proved that $EL = 2BL$ .

Step-by-step solution

Step 1: Prove congruence of triangles $\triangle EDM$ and $\triangle BCM$

Consider $\triangle EDM$ and $\triangle BCM$ . Since $AD \parallel BC$ and $AD$ is extended to $E$ , $AE \parallel BC$ , which gives alternate interior angles $\angle EDM = \angle BCM$ . Point $M$ is the midpoint of $CD$ , so $DM = CM$ , and vertically opposite angles are equal: $\angle EMD = \angle BMC$ . By the ASA congruence criterion, $\triangle EDM \cong \triangle BCM$ , which yields $ED = BC$ .

Step 2: Express the total length $AE$ in terms of $BC$

In parallelogram $ABCD$ , opposite sides are equal, so $AD = BC$ . From the previous congruence step, we found that $ED = BC$ . Adding these two segments gives the total length $AE = AD + ED = BC + BC = 2BC$ .

Step 3: Prove similarity of triangles $\triangle AEL$ and $\triangle CBL$

Now consider $\triangle AEL$ and $\triangle CBL$ . Since $AE \parallel BC$ , alternate interior angles give $\angle EAL = \angle BCL$ and $\angle AEL = \angle CBL$ . By the AA similarity criterion, $\triangle AEL \sim \triangle CBL$ . Therefore, the corresponding sides are proportional: $\frac{EL}{BL} = \frac{AE}{CB}$ .

Step 4: Substitute $AE = 2BC$ to conclude the proof

Substitute $AE = 2BC$ into the ratio from the similarity of triangles. This gives $\frac{EL}{BL} = \frac{2BC}{BC} = 2$ . Multiplying both sides by $BL$ yields the desired result: $EL = 2BL$ .

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Step-by-step solution

Step 1: Prove congruence of triangles △EDM\triangle EDM△EDM and △BCM\triangle BCM△BCM

Step 2: Express the total length AEAEAE in terms of BCBCBC

Step 3: Prove similarity of triangles △AEL\triangle AEL△AEL and △CBL\triangle CBL△CBL

Step 4: Substitute AE=2BCAE = 2BCAE=2BC to conclude the proof

Step 1: Prove congruence of triangles $\triangle EDM$ and $\triangle BCM$

Step 2: Express the total length $AE$ in terms of $BC$

Step 3: Prove similarity of triangles $\triangle AEL$ and $\triangle CBL$

Step 4: Substitute $AE = 2BC$ to conclude the proof