Question

in the given figure ba//ed . prove that angle abc + angle BCD = 180 + angle cde

Answer 1

Angle BCG + Angle ABC= 180 Angle DCG+ Angle CDE BCG + ABC + DCG= 180 + CDE Therefore, ABC + BCD= 180+ CDE.

Answer 2

Answer:

$\angle ABC+\angle BCD=180+\angle CDE$

Step-by-step explanation:

Given information: BA║ED

To prove: $\angle ABC+\angle BCD=180+\angle CDE$

Proof:

Draw a line PC, such that PC║BA and PC║ED.

If transversal line intersect two parallel lines, then alternate interior angle are same.

Since PC║ED, and DC is a transversal line, so

$\angle PCD=\angle CDE$                .... (1)

From the given figure it is clear that line PC divides the angle BCD in two parts.

$\angle BCP+\angle PCD=\angle BCD$

$\angle BCP=\angle BCD-\angle PCD$        .... (2)

If transversal line intersect two parallel lines, then the sum of  same sided interior angles is 180°.

Since PC║AB, and BC is a transversal line, so

$\angle ABC+\angle BCP=180$

Using equation (2) we get

$\angle ABC+(\angle BCD-\angle PCD)=180$             $[\becuase \angle BCP=\angle BCD-\angle PCD]$

Add $\angle PCD$ on both sides.

$\angle ABC+\angle BCD=180+\angle PCD$

Using equation (1) we get

$\angle ABC+\angle BCD=180+\angle CDE$            $[\becuase \angle PCD=\angle CDE]$

Hence proved.