Question

In a quad. ABCD, the line segments bisecting ZC and ZD meet at E. Prove that
ZA+ZB=2ZCED.​

Answer 1

∠1 = $\frac{1}{2}$, ∠2 $\frac{1}{2}$ ∠D

∆DEC

∠1 + ∠2 + ∠CED = 180°

∠CED = 180° - (∠1 + ∠2) -------------------- 1

∠A + ∠B + ∠C + ∠D = 360°

$\frac{1}{2}$ (∠A + ∠B) + $\frac{1}{2}$ ∠C + $\frac{1}{2}$ ∠D = 180°

$\frac{1}{2}$ (∠A + ∠B) + ∠1 + ∠2 = 180°

= $\frac{1}{2}$(∠A + ∠B) = 180° - (∠1 + ∠2) ------------------ 2

Equation 1 and 2

$\frac{1}{2}$ (∠A + ∠B) = ∠CED

∠A + ∠B = 2∠CDE