Geometry

[buoyant flower]

Given that the radius of the circumscribed circle of the quadrilateral ABCD is 5. Point E is the intersection of AC and BD. BE = DE, CD = √2DE, AC = 8. Calculate the area of ​​quadrilateral ABCD.

^ Please help me solve this question, I have spent way too much time on it and haven't succeeded.

[gloomy nymph]

△EDC and △CDB are similar, as CD/DE=BD/DC=√2 and they share a common angle.
Then △ADC is isosceles because ∠DAC=∠DBC=∠DCA. Thus, AC=8, sin(∠ADC)=4/5 and S(ABCD)=2S(ADC)=16.

[buoyant flower]

Hey, why is the area of ABCD twice the area of ADC?

[gloomy nymph]

ABCD consists of 2 triangles with common base and equal heights. One of them is ADC.

[buoyant flower]

Yes but im wondering how i can prove that their heights are equal.

[gloomy nymph]

BE=DE. Just project them onto AC.