The sidesBC,CA and AB of ΔABC are produced in order to form exterior angles ∠ACD,∠ABG, and ∠CAE then ∠ACD+∠CAE+∠ABG is
⇒ x+y+z=180∘
⇒ ∠ACD+∠CAE+∠ABG =(180∘−z)+(180∘−x)+(180∘−y)
⇒ ∠ACD+∠CAE+∠ABG =540∘−(x+y+z)=540∘−180∘=360∘
The sides AB, CA and AB of triangle ABCare produced in order to form exterior angles ACD, FBA,and angle CAE then angle ACD+CAE+ABF is
Sides BC, CA and AB of a triangle ABC are produced in an order, forming exterior angles ∠ACD, ∠BAE and ∠CBF. Show that ∠ACD + ∠BAE + ∠CBF = 360°.