∠ ECB+∠ BCA=180^∘ (Supplementary angles) 102 ^∘ #160; +∠ BCA= 180 ^∘ #160; ∠ BCA= 180 ^∘ #160; 102^∘ #160; ∠ BCA= 78 ^∘ #160; ...

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Corresponding Angles

∠ CED

Question

$∠ C E D$

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Solution

$∠ E C B + ∠ B C A = 180^{\circ}$ (Supplementary angles)
$102^{\circ} + ∠ B C A = 180^{\circ} ∠ B C A = 180^{\circ} - 102^{\circ} ∠ B C A = 78^{\circ}$
$∠ C E D = ∠ B C A$ (Corresponding angles as $B C | | E D$ and $C E$ is transversal)
$∠ C E D = 78^{\circ}$

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