A quadrilateral $A B C D$ is inscribed in a circle such that $A B: C D=2: 1$ and $B C: A D=5: 4$. If $A C$ and $B D$ intersect at the point $E$, then $A E$ : $C E$ equals

We have a quadrilateral ABCD inscribed in a circle with:

AB : CD = 2 : 1

BC : AD = 5 : 4

Diagonals AC and BD intersect at point E

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In a cyclic quadrilateral, when two angles are subtended by the same arc, they are equal.

From our diagram:

∠DAE and ∠CBE are both subtended by arc DC

∠ADE and ∠BCE are both subtended by arc AC

Therefore: ∠DAE = ∠CBE and ∠ADE = ∠BCE

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Since triangles AED and BEC have:

∠DAE = ∠CBE (angles subtended by same arc)

∠ADE = ∠BCE (angles subtended by same arc)

By AA similarity criterion, triangles AED and BEC are similar.

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From the similarity of triangles AED and BEC:

$\tfrac{AE}{BE} = \tfrac{AD}{BC} = \tfrac{DE}{CE}$

Using the given ratio BC : AD = 5 : 4, we get AD : BC = 4 : 5

Therefore: $\tfrac{AE}{BE} = \tfrac{4}{5}$ ... (1)

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Now let's look at triangles AEB and DEC:

∠BAE and ∠CDE are both subtended by arc BC

∠ABE and ∠DCE are both subtended by arc AD

So triangles AEB and DEC are also similar.

From this similarity:

$\tfrac{AE}{DE} = \tfrac{AB}{DC} = \tfrac{BE}{CE}$

Using the given ratio AB : CD = 2 : 1:

$\tfrac{AE}{DE} = \tfrac{2}{1} = 2$ ... (2)

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From equation (1): AE = $\tfrac{4}{5}$ × BE

From equation (2): AE = 2 × DE

Also from equation (2): BE = 2 × CE (since BE/CE = AB/DC = 2/1)

Now substituting:

From BE = 2 × CE, we get: AE = $\tfrac{4}{5}$ × (2 × CE) = $\tfrac{8}{5}$ × CE

Therefore: AE : CE = 8 : 5

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Answer: AE : CE = 8 : 5