i'm giving this a shot.. geometry is not my forte..

using property of similar triangles, AB/BC = AD/ DE

--> 4/6 = 6/DE

--> DE = 9

how do you know the triangles are similar?

money9111 wrote:how do you know the triangles are similar?

https://www.andrews.edu/~calkins/math/we ... 13.htm#SIM
Similar Triangles
If three sides of a triangle are proportional to the three sides of another triangle,
then the triangles are similar (SSS Similarity Theorem).

If the angles (two implies three) of two triangles are equal,
then the triangles are similar (AA Similarity Theorem).
Note: this applies not only to ASA, AAS=SAA, but also to AAA situations.

If two sides of a triangle are proportional to two sides of another triangle and the
included angles are congruent, then the triangles are similar (SAS Similarity Theorem).

Now, in this case the triangles in Question are ABC ande ADE

Angle AED = angle BED = angle DCB = Angle ACB (angles inscribed in the same arc and equal)
Angle ADE = angle CDE = angle CBE = Angle CBA (angles inscribed in the same arc and equal)

Now by AA theorem the triangles ABC ande ADE, are similar

I hope, I answered your question

skorolkova wrote:1)




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Angles in the same segment are equal. So, if BD is the chord, then ∠BCD = ∠BED; and if CE is the chord, ∠CBE = ∠CDE.

Also, ∠BAC = ∠DAE, as vertically opposite angles are equal.

With the help of above discussion, we can conclude that ∆ABC ~ ∆ADE, in that strict correspondence.

∴ AB/AD = BC/DE = CA/EA.

Let's take the needful only, i.e.

AB/AD = BC/DE

4/6 = 6/DE or DE = [spoiler]9[/spoiler].

money9111 wrote:how do you know the triangles are similar?

in tri ADE and ABC

<ade=<ABC ( ANGLE MADE BY TWO ON A CIRCLE ARE ALWAYS EQUAL)

FOR THE SAME REASON
<AED=<ACB

<DAE=<BAC(VERTICAL OPP ANGLES)

SO TRI ADE IS SIMILAR TO TRI ABC

THEREFORE AD/AB=DE/BC(PROP OF SIM TRI)

DE=6*6/4=9
HTH