Consider Delta DME and Delta CMB. Since AD BC, line AE is parallel to BC. With transversal CD, alternate interior angles EDM and BCM are equal. Vertically opposite angles DME and BMC are also equal. Given that M is the midpoint of CD, we have DM = MC. Therefore, by the ASA congruence criterion, Delta DME Delta CMB.

Consider Delta DME and Delta CMB. Since AD BC, line AE is parallel to BC. With transversal CD, alternate interior angles EDM and BCM are equal. Vertically opposite angles DME and BMC are also equal. Given that M is the midpoint of CD, we have DM = MC. Therefore, by the ASA congruence criterion, Delta DME Delta CMB.

Since corresponding sides of similar triangles are in proportion, we have (EL)/(BL) = (AE)/(CB). Substituting AE = 2BC into the ratio gives (EL)/(BL) = (2BC)/(BC) = 2. Multiplying both sides by BL gives the required relation EL = 2BL.

Since corresponding sides of similar triangles are in proportion, we have (EL)/(BL) = (AE)/(CB). Substituting AE = 2BC into the ratio gives (EL)/(BL) = (2BC)/(BC) = 2. Multiplying both sides by BL gives the required relation EL = 2BL.

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

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

Step-by-step solution

Step 1: Prove congruence of triangles DME and CMB

Consider $\Delta DME$ and $\Delta CMB$ . Since $AD \parallel BC$ , line $AE$ is parallel to $BC$ . With transversal $CD$ , alternate interior angles $\angle EDM$ and $\angle BCM$ are equal. Vertically opposite angles $\angle DME$ and $\angle BMC$ are also equal. Given that $M$ is the midpoint of $CD$ , we have $DM = MC$ . Therefore, by the ASA congruence criterion, $\Delta DME \cong \Delta CMB$ .

Step 2: Relate segment AE to BC

From the congruence of $\Delta DME$ and $\Delta CMB$ , corresponding parts are equal, so $DE = BC$ . In parallelogram $ABCD$ , opposite sides are equal, which means $AD = BC$ . Therefore, the total segment $AE$ is given by $AE = AD + DE = BC + BC = 2BC$ .

Step 3: Establish similarity between triangles AEL and CBL

In $\Delta AEL$ and $\Delta CBL$ , since $AE \parallel BC$ , alternate interior angles $\angle EAL$ and $\angle BCL$ are equal. Additionally, vertically opposite angles $\angle ALE$ and $\angle CLB$ are equal. Hence, by the AA similarity criterion, $\Delta AEL \sim \Delta CBL$ .

Step 4: Conclude that EL = 2BL

Since corresponding sides of similar triangles are in proportion, we have $\frac{EL}{BL} = \frac{AE}{CB}$ . Substituting $AE = 2BC$ into the ratio gives $\frac{EL}{BL} = \frac{2BC}{BC} = 2$ . Multiplying both sides by $BL$ gives the required relation $EL = 2BL$ .

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