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Q.
In the given figure, $AB \| CD , \angle EAB =30^{\circ}$ and $\angle EDC =40^{\circ}$. Then the measure of $\angle DEA$ is_____
Geometry
Solution:
Given $AB \| CD$
Draw a line (l) passing through E, parallel to $AB$.
Let us say $l$ meets $AD$ at $P$.
$\therefore AB \|l \Rightarrow CD \| I $
$\angle AEP =\angle EDC =40^{\circ} \text { ( } \because \text { Alternate angles) } $
$\angle PEA =\angle EAB =30^{\circ} \text { ( } \because \text { Alternate angles) } $
$\therefore \angle DEA =\angle DEP +\angle PEA =40^{\circ}+30^{\circ} $
$\angle DEA =70^{\circ}$
