geometry problem: "Consider a semicircle where AB is the diameter. On this semicircle, there's a point C..."

[askingme]

Hi folks, I have a geometry problem that I have stuck with for a while.
Can anyone help me?
Thanks a lot

[problem]
Consider a semicircle where AB is the diameter. On this semicircle, there's a point C, and D is located at the midpoint of the smaller arc BC. The lines AD and BC cross at point E. Given that the length of CE is 3 units and the length of BD is 2√5 units, determine the length of the diameter AB.

 

[stapel]

Hi folks, I have a geometry problem that I have stuck with for a while.
Can anyone help me?

Consider a semicircle where AB is the diameter. On this semicircle, there's a point C, and D is located at the midpoint of the smaller arc BC. The lines AD and BC cross at point E. Given that the length of CE is 3 units and the length of BD is 2√5 units, determine the length of the diameter AB.

We'll be glad to help! But first we need to see what you've tried and how far you've gotten. ("Read Before Posting")

Please reply with a clear listing of your thoughts and efforts for at least one of the attempts that you've made during the "while" for which you've been working on this. Thank you!

 

[blamocur]

Making a drawing might be a good first step.

 

[Harry_the_cat]

Just checking that BD is 2√5 units, not BE?

 

[Dr.Peterson]

Hi folks, I have a geometry problem that I have stuck with for a while.
Can anyone help me?
Thanks a lot

[problem]
Consider a semicircle where AB is the diameter. On this semicircle, there's a point C, and D is located at the midpoint of the smaller arc BC. The lines AD and BC cross at point E. Given that the length of CE is 3 units and the length of BD is 2√5 units, determine the length of the diameter AB.

I'd start by drawing a picture and labeling any segments I could with their lengths. I'd also look for similar triangles and right triangles.

Then I'd show my picture to the people I was asking for help.

 

[pka]

[problem]
Consider a semicircle where AB is the diameter. On this semicircle, there's a point C, and D is located at the midpoint of the smaller arc BC. The lines AD and BC cross at point E. Given that the length of CE is 3 units and the length of BD is 2√5 units, determine the length of the diameter AB.

Hints:
#1 You have a midpoint, arcs of the same measure have chords of the same measure,
#2 Can you show that [imath]\Delta AEB \approx \Delta CED[/imath] ?
#3 Pay close attention to that of corresponding vertices.

 

[askingme]

Hints:
#1 You have a midpoint, arcs of the same measure have chords of the same measure,
#2 Can you show that [imath]\Delta AEB \approx \Delta CED[/imath] ?
#3 Pay close attention to that of corresponding vertices.

Thanks for this.
1 yes, i have find the CD=BD
2 yes, this two triangles are similar to each other and i can have 2√5/2r = 3/AE
3 i have no idea how to properly use CD=BD

Can i have more hints please?

 

[askingme]

We'll be glad to help! But first we need to see what you've tried and how far you've gotten. ("Read Before Posting")

Please reply with a clear listing of your thoughts and efforts for at least one of the attempts that you've made during the "while" for which you've been working on this. Thank you!

i can tell CD=BD
and i want to use ptolemy to have CD2r + DBAC=ADCB
also i want to try pY in triangle ADB and EDB, but both involves heavy calculation.
i am thinking about some quick way to solve the problem...

 

[askingme]

i think my approach involves too much calculation and wondering if there is anything easier?

 

[limiTS]

One or more of the lengths is wrong, either CE or BD or both. As written, the ratio [imath]\dfrac{\text{BD}}{\text{CE}}=\dfrac{2\sqrt{5}}{3}\approx1.49[/imath], and this value stays the same as the radius is scaled up or down and point C moves proportionally.

However, the maximum ratio is [imath]\dfrac{\text{BD}}{\text{CE}}\approx1.31[/imath], and it occurs when C is directly over the semicircle's center.

Instead, I solved it for [imath]\dfrac{\text{BD}}{\text{CE}}=\dfrac{2\sqrt{5}}{\color{red}{4}}\approx1.12[/imath], and here's the construction in Desmos. Grid numbers were removed to not give away the answer.

 

[BobP]

One or more of the lengths is wrong, either CE or BD or both. As written, the ratio [imath]\dfrac{\text{BD}}{\text{CE}}=\dfrac{2\sqrt{5}}{3}\approx1.49[/imath], and this value stays the same as the radius is scaled up or down and point C moves proportionally.

However, the maximum ratio is [imath]\dfrac{\text{BD}}{\text{CE}}\approx1.31[/imath], and it occurs when C is directly over the semicircle's center.

Instead, I solved it for [imath]\dfrac{\text{BD}}{\text{CE}}=\dfrac{2\sqrt{5}}{\color{red}{4}}\approx1.12[/imath], and here's the construction in Desmos. Grid numbers were removed to not give away the answer.


View attachment 36784

There is nothing wrong with the question as it is.
There is no reason why C can't be closer to A than it is to B, (so that your ratio 1.49 is attained).
The 'minor arc BC ', is simply saying that D lies on the arc BC that doesn't contain A.

As to a method of solution, begin by adding the lines AC, OC and OD, where O is the centre of the circle.
Notice that since D is at the mid-point of the arc BC, the angles COD and BOD will be equal.
Call one of those angles 2theta, say, (the 2 avoids fractions later), and fill in a few more angles, (in terms of theta), around the diagram.
Now using the fact that the triangles ACE and ADB are right-angled, derive two equations in theta and r, the radius of the circle.
Solve them simultaneously.

 

[limiTS]

I interpreted the OP's "smaller arc BC" to mean smaller than AC. But if you go with BobP's interpretation, meaning C can be left of the center, then [imath]\text{BD}=2\sqrt{5}[/imath] and [imath]\text{CE}=3[/imath] does yield a simple solution. Here is the corrected graph (without giving it away):