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Hi there, the following solution should be the standard algebraic method.

∠CAE = ∠DAE (Straight Line)

∠DAE = ∠DEA (Isos Triangle Properties)

∠ADE = 180 – ∠DAE – ∠DEA (∠s in a Triangle)

∠ADE = 180 – 2∠CAE ———– (1)

Focusing on Triangle CDO

∠CAE = ∠DCO (Isos Triangle Properties)

∠CDO = 180 – 54 – ∠DCO (∠s in a Triangle)

∠CDO = 126 – ∠CAE ——— (2)

Focusing on the Straight Line ADC

∠ADE = 90 – ∠CDO (∠s on Straight Line ADC)

Substituting (1) and (2) into the equation above

(180 – 2∠CAE) =  90 –  (126 – ∠CAE)

216 = 3∠CAE

Answer: ∠CAE = 72

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