Question

if $\angle ecd = 45^{\circ}$ and $\angle bed = 110^{\circ}$ is $\triangle abe \sim \triangle cde$? if so, by what criterion?
a yes, by aa criterion
b yes, by sas criterion
c yes, by ssa criterion
d no, not possible to tell.

Explanation:

1. First, find \( \angle ABE \) and \( \angle CED \):

• Given \( \angle BED = 110^\circ \), so \( \angle CED=180^\circ - 110^\circ = 70^\circ \) (linear pair).
• In \( \triangle CDE \), \( \angle ECD = 45^\circ \), \( \angle CED = 70^\circ \), so \( \angle D=180^\circ-(45^\circ + 70^\circ)=65^\circ \).
• We know \( \angle ABE = 65^\circ \) (given) and \( \angle D = 65^\circ \), so \( \angle ABE=\angle D \).
• Also, \( \angle BAE \) and \( \angle DCE \): Wait, actually, \( \angle AEB \) and \( \angle CED \)? No, wait, \( \angle A=\angle ECD \)? Wait, no, let's re - check. Wait, \( \angle ABE = 65^\circ \), \( \angle D = 65^\circ \) (so one pair of equal angles). Then, \( \angle AEB \): Wait, \( \angle BED = 110^\circ \), so \( \angle AEB = 180^\circ - 110^\circ=70^\circ \), and \( \angle CED = 70^\circ \) (from earlier). So \( \angle AEB=\angle CED \). So now we have two pairs of equal angles: \( \angle ABE=\angle D = 65^\circ \) and \( \angle AEB=\angle CED = 70^\circ \). By AA (Angle - Angle) similarity criterion, if two angles of one triangle are equal to two angles of another triangle, the triangles are similar. So \( \triangle ABE\sim\triangle CDE \) by AA criterion.

Answer:

A. yes, by AA criterion