In EDM and BCM, since line AD is parallel to BC, line AE is parallel to BC. Therefore, the alternate interior angles satisfy EDM = BCM and DEM = CBM. Since M is the midpoint of CD, we have DM = MC. Also, the vertically opposite angles D M E and C M B are equal. Hence, by ASA congruence (or AAS congruence), EDM BCM, which gives ED = BC by CPCTC.

In EDM and BCM, since line AD is parallel to BC, line AE is parallel to BC. Therefore, the alternate interior angles satisfy EDM = BCM and DEM = CBM. Since M is the midpoint of CD, we have DM = MC. Also, the vertically opposite angles D M E and C M B are equal. Hence, by ASA congruence (or AAS congruence), EDM BCM, which gives ED = BC by CPCTC.

Because ABCD is a parallelogram, its opposite sides are equal, so AD = BC. From the congruence of EDM and BCM, we established that ED = BC. Adding these two segments together gives AE = AD + ED = BC + BC = 2BC.

Because ABCD is a parallelogram, its opposite sides are equal, so AD = BC. From the congruence of EDM and BCM, we established that ED = BC. Adding these two segments together gives AE = AD + ED = BC + BC = 2BC.

In AEL and CBL, the lines AE and BC are parallel, cut by transversals AC and BE. This gives alternate interior angles EAL = BCL and AEL = CBL. Moreover, the vertically opposite angles ALE and CLB are equal. Therefore, by AA similarity, AEL CBL. The ratio of their corresponding sides is thus (EL)/(BL) = (AE)/(BC).

In AEL and CBL, the lines AE and BC are parallel, cut by transversals AC and BE. This gives alternate interior angles EAL = BCL and AEL = CBL. Moreover, the vertically opposite angles ALE and CLB are equal. Therefore, by AA similarity, AEL CBL. The ratio of their corresponding sides is thus (EL)/(BL) = (AE)/(BC).

Substitute AE = 2BC into the ratio of corresponding sides (EL)/(BL) = (AE)/(BC). The side length BC cancels out, yielding (EL)/(BL) = 2. Cross-multiplying gives the desired result EL = 2BL.

$ABCD$ is a parallelogram ( $||gm$ ). $BE$ bisects $CD$ at $M$ and intersects $AC$ at $L$ . Prove that $EL = 2 BL$ . Figure Details: - A parallelogram $ABCD$ is drawn with bottom vertices $A$ (left), $B$ (right), and top vertices $D$ (left), $C$ (right). - Side $AD$ is extended upwards to a point $E$ . - Line segment $BE$ is drawn, intersecting the side $CD$ at point $M$ and the diagonal $AC$ at point $L$ . - The diagonal $AC$ is drawn, connecting vertex $A$ to vertex $C$ . - $M$ is the midpoint of $CD$ (as $BE$ bisects $CD$ at $M$ ).

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

Step-by-step solution

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

In $\triangle EDM$ and $\triangle BCM$ , since line $AD$ is parallel to $BC$ , line $AE$ is parallel to $BC$ . Therefore, the alternate interior angles satisfy $\angle EDM = \angle BCM$ and $\angle DEM = \angle CBM$ . Since $M$ is the midpoint of $CD$ , we have $DM = MC$ . Also, the vertically opposite angles $\angle D M E$ and $\angle C M B$ are equal. Hence, by ASA congruence (or AAS congruence), $\triangle EDM \cong \triangle BCM$ , which gives $ED = BC$ by CPCTC.

Step 2: Express $AE$ in terms of $BC$

Because $ABCD$ is a parallelogram, its opposite sides are equal, so $AD = BC$ . From the congruence of $\triangle EDM$ and $\triangle BCM$ , we established that $ED = BC$ . Adding these two segments together gives $AE = AD + ED = BC + BC = 2BC$ .

Step 3: Establish similarity between $\triangle AEL$ and $\triangle CBL$

In $\triangle AEL$ and $\triangle CBL$ , the lines $AE$ and $BC$ are parallel, cut by transversals $AC$ and $BE$ . This gives alternate interior angles $\angle EAL = \angle BCL$ and $\angle AEL = \angle CBL$ . Moreover, the vertically opposite angles $\angle ALE$ and $\angle CLB$ are equal. Therefore, by AA similarity, $\triangle AEL \sim \triangle CBL$ . The ratio of their corresponding sides is thus $\frac{EL}{BL} = \frac{AE}{BC}$ .

Step 4: Substitute and conclude the proof

Substitute $AE = 2BC$ into the ratio of corresponding sides $\frac{EL}{BL} = \frac{AE}{BC}$ . The side length $BC$ cancels out, yielding $\frac{EL}{BL} = 2$ . Cross-multiplying gives the desired result $EL = 2BL$ .

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Step-by-step solution

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

Step 2: Express AEAEAE in terms of BCBCBC

Step 3: Establish similarity between △AEL\triangle AEL△AEL and △CBL\triangle CBL△CBL

Step 4: Substitute and conclude the proof

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

Step 2: Express $AE$ in terms of $BC$

Step 3: Establish similarity between $\triangle AEL$ and $\triangle CBL$