Question

in the given figure AB parallel to CD prove that BAE-DCE=AEC

Answer 1

here is your answer....

Answer 2

The proof of $\angle BAE- \angle DCE =\angle AEC$ is shown below.

Step-by-step explanation:

We are given the figure in the question in which AB║CD.

Firstly, we will make some construction in the figure such that;

Draw a line EF through point E which is parallel to the line AB and CD.

This means EF║AB ║CD.

Since EF║AB and AE act as a transversal, this means that;

$\angle$AEF and $\angle$BAE are supplementary because the sum of co-interior angles is 180°, i.e;

$\angle$AEF + $\angle$BAE = 180°  

$\angle$AEF = 180° -  $\angle$BAE     ----------------- [Equation 1]

Similarly, EF║CD and CE act as a transversal, this means that;

$\angle$CEF and $\angle$DCE are supplementary because the sum of co-interior angles is 180°, i.e;

$\angle$CEF + $\angle$DCE = 180°   ----------------- [Equation 2]

Now, from the diagram it is clear that $\angle$CEF can be written as a sum of $\angle$AEF and $\angle$AEC, that means;

$\angle$CEF + $\angle$DCE = 180°  

($\angle$AEF +  $\angle$AEC) + $\angle$DCE = 180°  

180° -  $\angle$BAE +  $\angle$AEC + $\angle$DCE = 180°    {using equation 1}

    $\angle$BAE -  $\angle$DCE =  $\angle$AEC    {Hence Proved}