In a quad. ABCD, the line segments bisecting ZC and ZD meet at E. Prove that
ZA+ZB=2ZCED.
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∠1 = , ∠2 ∠D
∆DEC
∠1 + ∠2 + ∠CED = 180°
∠CED = 180° - (∠1 + ∠2) -------------------- 1
∠A + ∠B + ∠C + ∠D = 360°
(∠A + ∠B) + ∠C + ∠D = 180°
(∠A + ∠B) + ∠1 + ∠2 = 180°
= (∠A + ∠B) = 180° - (∠1 + ∠2) ------------------ 2
Equation 1 and 2
(∠A + ∠B) = ∠CED
∠A + ∠B = 2∠CDE
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