Question

The figure shows two parallel lines AB and DE cut by the transversals AE and BD: AB and DE are parallel lines and AE and BD are transversals. The transversals intersect at C. Angle CAB is labeled 1, angle ABC is labeled 2, angle ACB is labeled 3, angle ACE is labeled 4, angle CDE is labeled 6, and angle CED is labeled 5. Which statement best explains the relationship between Triangle ABC and Triangle EDC?
Triangle ABC is similar to triangle EDC because Measure of angle 3 equals measure of angle 6 and measure of angle 1 equals measure of angle 4

Triangle ABC is similar to triangle EDC because Measure of angle 3 equals measure of angle 4 and Measure of angle 1 equals measure of angle 5

Triangle ABC is congruent to triangle EDC because Measure of angle 3 equals measure of angle 4. and measure of angle 1 equals measure of angle 5.

Triangle ABC is congruent to triangle EDC because Measure of angle 3 equals measure of angle 6. and Measure of angle 1 equals measure of angle 4.

Answer 1

Answer: Triangle ABC is similar to triangle EDC because Measure of angle 3 equals measure of angle 4 and Measure of angle 1 equals measure of angle 5

Step-by-step explanation:

Given: $AB \parallel DE$,  $AE\cap BD = C$

Where AE and BD are the common transversals of parallel lines AB and DE,

Also, Angle CAB is labeled 1, angle ABC is labeled 2, angle ACB is labeled 3, angle DCE is labeled 4, angle CDE is labeled 6, and angle CED is labeled 5.

Thus,   $\angle 3 \cong \angle 4$   ( Vertically opposite angles )

$\angle 1 \cong \angle 5$            ( Alternative interior angle theorem)

Thus, By AA similarity postulate,

$\triangle ABC \sim \triangle EDC$

Hence, proved.

Second Option is correct.