Question

Show that $\angle ACD+\angle BAE+\angle CBF={360}^{\circ }$

Answer 1

Show that, $\angle ACD+\angle BAE+\angle CBF={360}^0$

Proof: In the given figure

$\angle ACD=\angle ABC+\angle BAC.........\left(1\right)$

$\angle BAE=\angle ABC+\angle ACB..........\left(2\right)$

$\angle CBF=\angle BAC+\angle ACB...........\left(3\right)$

On adding (1), (2) and (3) to, and we get,

$\angle ACD+\angle BAE+\angle CBF=2\angle BAC+2\angle ABC+2\angle ACB$

$\angle ACD+\angle BAE+\angle CBF=2(\angle BAC+\angle ABC+\angle ACB)$

Using triangle angle sum formula,

$\angle ACD+\angle BAE+\angle CBF=2\times {180}^0$

$\angle ACD+\angle BAE+\angle CBF={360}^0$

Hence proved.