Question

In the figure prove that <BAD = <DCE​

Answer 1

Step-by-step explanation:

$Produce \: BA \: to \: touch \: EC. Let \: the \: point \: of \: intersection \: be \: F.$

intersection be F.

$∠AFE=∠ECD$

$Also, ∠BAE=∠AFE+∠AEC</p><p></p><p> \: \: Therefore,</p><p></p><p>$

$∠BAE=∠ECD+∠AEC</p><p></p><p>∠BAE−∠ECD=∠AEC</p><p></p><p>$