Prove that ##AE=2BC## -Deductive Geometry

[chwala]

@anuttarasammyak , attached is the same drawing, with improved resolution for the angles and measurements of the sides of the triangles.
View attachment 315404

Nice ...what's your ultimate goal/intention on having different drawings on the problem? are the measurements of $AE$ and$BC$ tending towards the envisaged value? Or drifting away...cheers.

 

[SammyS]

Looking at it again...in my opinion i tend to think the concept being tested on this question is congruency. In answering this question it is paramount to use a given circle property.

. . .

Clearly, the textbook has a mistake! The correct statement should have been

Prove $AB=2BC$.

I agree with @anuttarasammyak .

The textbook problem is solvable.

As he said, you simply need to show that $\displaystyle \triangle DCA \sim \triangle BEA$ . From this you get that $AE=2EB$ .

Also, point $O$ is not necessarily on $BD$ .

 

[chwala]

I agree with @anuttarasammyak .

The textbook problem is solvable.

As he said, you simply need to show that $\displaystyle \triangle DCA \sim \triangle BEA$ . From this you get that $AE=2EB$ .

Also, point $O$ is not necessarily on $BD$ .

Kindly find my rejoinder...sorry am a bit busy had to scribble my response.
I have made use of similarity (on the diagram) as is suggested...and I still cannot see how $BC= EB$.

 

[anuttarasammyak]

There is no mention in the problem that $\angle ACD$ is rectangle or BD is diameter. BD can be a diameter as @Lnewqban showed. But it is not a requirement coming from AC = 2CD that BD is a diameter.

 

[chwala]

View attachment 315462
There is no mention in the problem that $\angle ACD$ is rectangle or BD is diameter. AC = 2CD does not require that BD is a diameter.

If indeed it's not a diameter then we cannot talk of Similarity. That would mean that we are looking at the other option of Congruency. In that case, it would be interesting to see which postulate applies.

 

[anuttarasammyak]

What similarity are you worrying about ?

As
BC=BE
We should prove
\triangle DCA \sim \triangle BEA

This similarity holds with no regard to BD is a diameter or not.

 

[chwala]

What similarity are you worrying about ?This similarity holds with no regard to BD is a diameter or not.

Ok then, can you show how $BC=BE$? Can you draw a sketch of your diagrams separately for clarity.

 

[chwala]

I do not need BC=BE for the similarity. Instead, I used equality of two corresponding angles.

This means you're not in agreement with post $30$. Does the diagram depict the problem? I used it in my analysis.
Is $BE$ equal to $BC$ as indicated?if so, is my analysis correct as shown on post $33$?

 

[anuttarasammyak]

I think @Lnewqban showed us a nice drawing solving the problem with an additional condition of "BD is a diameter".

 

[chwala]

Can we share the idea that two triangle are Similar if two corresponding angles are same ?

A sketch of your similar triangles would be adequate. Thanks.

 

[anuttarasammyak]

Is $BE$ equal to $BC$ as indicated?if so, is my analysis correct as shown on post $33$?

Arc BE and Arc BC have same angle of circumference or central angle. Their length are equal thus BE=BC.

I apologize saying "I do not need BE=BC" and "congruence and similarity" in error.

 

[Lnewqban]

Nice ...what's your ultimate goal/intention on having different drawings on the problem?

No specific goal or intention.
Just trying to understand the problem, which results interesting to me.
The first drawing was the result of beginning with assumed angles: it did not work.
The second drawing was the result of beginning with the information given in the text of the problem: it worked.
The third drawing is exactly as the second one, only showing the value of the resulting angles with a resolution of 0.00, so @anuttarasammyak could see that those values correspond to the 1:2 proportion of the sides of the triangles.

... are the measurements of $AE$ and$BC$ tending towards the envisaged value? Or drifting away...cheers.

AE (1.24 units) is twice as long as BE (0.62 units).
The discontinuos-line circle shows us that the lenghts of BE and BC are exactly the same.

 

[Lnewqban]

I understand that the graphic way is not the way for any student to prove what the problem requires to be proved, as you stated before.
The problem is that I don't know enough math or trigonometry to prove it mathematically.
I hope that my drawings help you gentlemen to find a way to do so, as they prove that the problem is properly stated and that is solvable.

 

[chwala]

I understand that the graphic way is not the way for any student to prove what the problem requires to be proved, as you stated before.
The problem is that I don't know enough math or trigonometry to prove it mathematically.
I hope that my drawings help you gentlemen to find a way to do so, as they prove that the problem is properly stated and that is solvable.

I appreciate your input @Lnewqban ...i would appreciate it if you may have/draw the two similar triangles side by side. That is what i asked for in post $40$...

or alternatively,

tell me what is wrong in my sketch i.e post $33$ as the analysis is directly drawn from your graph drawing of your post $30$.
Cheers bro.

 

[chwala]

@anuttarasammyak , attached is the same drawing, with improved resolution for the angles and measurements of the sides of the triangles.
View attachment 315404

Your diagram looks accurate and convincing! Just one question, i thought for similar Triangles we need to have two pair of corresponding angles which is not the case in your diagram...

you have $90^0, 26.57^0, 63.43^0$ on one triangle and $90^0, 31.72^0,58.28^0$ on the other triangle.

Otherwise, if we were to go with your diagram, then the assertion $AE=2BC$ is correct.

 

[anuttarasammyak]

Please find below my rough hand drawing of the problem. It happens to be the case of
AC/CD = AE/BC = about 0.8. O is outside DE.
May I understand that you say the ratio 2 is the special case that O must be on BD ?

 

[chwala]

Please find below my rough hand drawing of the problem. It happens to be the case of
AC/CD = AE/BC = about 0.8. O is outside DE.
May I understand that you say the ratio 2 is the special case that O must be on BD ?
View attachment 315472

nice @anuttarasammyak ... looks like after all... $AE≠2BC$...rather;

From your analysis; $AE=\dfrac{4}{5}BC$

 

[Steve4Physics]

May I chip in.

Although the diagram makes BD look like it might be a diameter, there is no reason to assume it is (as I think has already been noted).

Consequently, we can’t assume things like ∠EBD = ∠CBD or that ∠BCD = 90º.

There is nothing wrong with the problem statement.

If you are familiar with certain theorems, the solution is straightforward.

The intersecting secants theorem tells us that:
AB•AC = AE•AD.

The angle bisector theorem tells us that:
AD•BC= CD•AB

We are given AC = 2CD.

Proving that AE=2BC requires only simple algebra using the above equations.

 

[anuttarasammyak]

The essence of the problem is
AC/CD=AE/BC
coming from same arc lengths and similarity of triangles.

I draw another case which happens to have the ratio of about 2.4 and a procedure to find the case of ratio 2. With fixed B and D, O keeps out of BD.

 

[Steve4Physics]

FWIW, I agree with @anuttarasammyak.

BD does not need to be a diameter for the 'similarity' approach to be valid.

This follows from the "Angles of Intersecting Chords Theorem". E.g. see https://www.varsitytutors.com/hotmath/hotmath_help/topics/angle-of-intersecting-chords-theorem

The arc length depends on the angle between the chords.

The angle between DC and DB is the same as the angle between DE and DB. (With E being the point of ‘intersection’ of these 3 chords). If follows that arc length BC = arc length BE.

 

[SammyS]

Kindly find my rejoinder...sorry am a bit busy had to scribble my response.
I have made use of similarity (on the diagram) as is suggested...and I still cannot see how $BC= EB$.

That $BC=EB$ does not come from similarity of triangles. It comes from the fact that chord $BC$ and chord $EB$ each subtend inscribed angles of equal measure, namely $\angle BDC$ and $\angle EDB$ .

Steve beat me to it !.

 

[anuttarasammyak]

The essence of the problem is
AC/CD=AE/BC
coming from same arc lengths and similarity of triangles.

Or $\triangle EOB \equiv \triangle COB$ instead of same arc lengths, if you prefer.

 

[chwala]

...just going through my posts...this was a nice one...just to remember how i understand it in the near future; i reflected $AE$ on the supposedly mirror line $AC$, it then follows that;

$\dfrac{AE}{BE}=\dfrac{AC}{CD}$

$\dfrac{AE}{BC}=\dfrac{AC}{CD}$

We are given; $AC=2CD$

$\dfrac{AE}{BC}=\dfrac{2CD}{CD}$

$\dfrac{AE}{BC}=2$

$⇒AE=2BC$.