Following are the response to the given equation:
Given:
[tex]\bold{ m \angle ABC=53^{\circ}}\\\\\bold{m \angle ECD=37^{\circ}}\\\\[/tex]
To find:
[tex]\bold{\Delta BAC \sim \Delta EDC=?}[/tex]
Solution:
[tex]\bold{\angle ABC=53^{\circ}}\\\\\bold{\angle ECD=37^{\circ}}\\\\[/tex]
If
[tex]\to \bold{\angle ECD=37^{\circ}}\ \ \ \ \ \ \ \ \ \ then \ \ \ \ \ \ \ \ \ \bold{\angle CED=(90-37)^{\circ}= 53^{\circ}}\\\\[/tex]
therefore
[tex]\bold{\angle ABC =\angle CED }\\\\[/tex]
Yes, because the AA similarity postulate could be applied since a reflection over line f proves that [tex]\bold{\angle ABC \cong \angle DEC}[/tex] .
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