In the circle inscribed quadrilateral ABCD, AC and BD intersect at point E, and AE = CE, prove ad × AB = DC × BC

In the circle inscribed quadrilateral ABCD, AC and BD intersect at point E, and AE = CE, prove ad × AB = DC × BC

It is proved that: ∵ - BAE = CDE, ∵ - AEB = Dec
∴△ABE≈△DCE
∴AB/CD=BE/CE
Similarly, △ ade ≈ △ BCE
∴AD/BC=AE/BE
∴(AB/CD)×(AD/BC)=AE/CE=1
∴AD×AB=DC×BC

As shown in the figure, in the circle inscribed quadrilateral ABCD, AC and BD intersect at point E, and AE = CE, proving: ad.ab = dc.bc

Because: AE = EC, the circle is inscribed with quadrilateral ABCD
So: s △ abd = s △ ACB, sin ∠ a = sin ∠ C
Because: s △ abd = 1 / 2ad * AB * Sina, s △ ACB = 1 / 2dc * BC * sinc
So: 1 / 2ad * AB * Sina = 1 / 2dc * BC * sinc
So: ad * AB = DC * BC

Given that the quadrilateral ABCD is inscribed in a circle, ad is the diameter, ad = 8, ab = BC, CD = 7, then what is ab equal to?

Even AC and Bo are handed over to E,
In ACD, by Pythagorean theorem, AC = √ (AD ^ 2-CD ^ 2) = √ 15,
Because AB = BC,
So AE = AC / 2 = √ 15 / 2,
In right triangle AOE, by Pythagorean theorem, OE = √ (AO ^ 2-ae ^ 2) = 7 / 2,
So be = bo-eo = 4-7 / 2 = 1 / 2,
In the right triangle Abe, by Pythagorean theorem, ab = √ (AE ^ 2-BE ^ 2) = 2