Find BC:CD

[ttokoro]

Find BC:CD

[AlfredFlaßhaar · Accepted solution]

A memory from school days :-):
Since only the aspect ratio is sought, the choice of the line lengths denoted by the ratio 4/7 is irrelevant – this is a problem of so-called similarity geometry. In all situations similar to the given geometric structure, the same ratio results in the solution. Therefore, with the help of a calculator, paper, and pen, choose AC and AD in the assumed ratio ---> BC/CD = 1.333...
Perhaps there is a clever solution using reflection/symmetry that doesn't require any calculations. I'll have to puzzle it out.

[Werner_E]

I can confirm your result, but unfortunately Mathcads symbolics is not able to simplify it to 4/3 which probably is the exact result.

And I am pretty sure that @ttokoro  will provide a very basic and clever geometric solution without much/any calculations 😉

Her my brute force attack in MC15

[Werner_E]

Let's consider the angle symmetry w of angle CAD.

Because 77° + 13° = 90°, it is at right angles to AB.

Now, if we reflect point C across w, we get point E on the line segment DA.

CE is therefore parallel to AB (both are perpendicular to w).

Now draw a parallel to BC through point E and intersect it with BA at point F.

If we choose AC=4 and AD=7, then AE=4 and ED=3.

The triangles CDE and FEA are similar triangles, so because FE=BC, we get the ratio

FE:CD = BC:CD=AE:ED=4:3

//.

[Werner_E]

Ooops, OK, a much easier and more straightforward way is using the intercept theorem.

Lets construct point E as described above which intersects the line AD at the ratio 4:3.

Because AB and CE are parallel it follows immediately that BC:CD = 4:3

[SPaulis]

I think if you catch that angel BAC + half of angle CAD = 90°, like @WernerE did, an answer without brute force becomes apparent pretty quick.  I did not catch it myself—I tried the brute force with sine laws and got 1.33333.  I could not simplify my equation to something elegant.

[Werner_E]

Looks like old Mathcad 15 with muPad is a bit more capable here

Prime 10.0.1.0:

Maybe  applying the special relationship between the two angles can help ...

Disappointment!

So lets manually apply some double/half angle rules

Heureka! Now even Prime is able to finish the job!

It even works symbolically without assigning values for CAD and AC:AD

So it seems that Prime's Symbolic doesn't know all that much, but it obviously at least knows the relationship

and therefore can simplify

Nonetheless its much simpler and straightforward in real Mathcad

But one should not give up hope - perhaps one day Prime will also grow up and be more or less on a par with good old Mathcad. After all it's been less than 20 years since Prime 'replaced' Mathcad as the successor. 😉

[SPaulis]

@Werner_E but there is no reason to simply to your last ratio, although elegant.  This solution is dependent on enough input geometry so that simple sine laws are sufficient.  The uniqueness of this problem is to catch what you did originally, that " angel BAC + half of angle CAD = 90° ", so that you didn't need to do any math.  If anyone changed the input parameter angle BAC (i.e. length BC), it would be a simple high school geometry problem to practice the sine law.  The issue I have is that this problem doesn't really promote the effectiveness of Mathcad because you came up with the 'elegant' answer in like... 10 seconds!  This is more like an IQ testing question where you'd fail if you 'had to' pull out your calculator. 

You had me at "Let's consider..." 😉

[Werner_E]

I still feel that its necessary for a complete solution to get the exact result 4/3 rather than the numeric 1,3333 one. Or in case of the generic solution its important to see that the result is a simple rational number (assuming that CD:AD is rational) and not an irrational value derived by sine functions.

Unfortunately Prime needed a lot of help to arrive at the exact result.

Concerning spotting the 77°+1/2*26° = 90° - adding or subtracting angles  very often is the key to the solution with puzzles like this where seemingly strange and random angles are used.
I also vaguely remember that @ttokoro already posted a similar puzzle some time ago where the key to success also was to see that some of the given angles add up to or make a difference of a 'nice' angle like 90° or 180°.
Not sure if I have this one in mind Solved: Find BH:HC - PTC Community or some other puzzle he posted.

@ttokoro now posted here a picture where he used that 2*77°+26° = 180°. I noticed that, too (actually that lead me to the "half" relationship which I finally used) but could not find any use for it.

I also can't see the solution in the picture posted by @ttokoro .

[ttokoro]

Thanks all how to solve this puzzle by Mathcad.

[Werner_E]

I see that D, A and E are collinear and that BC=BE. Basically ABC is mirrored at AB.

But I can't see the answer BC:CD=4:3 in your picture ....???

Of course, it's quite possible that I'm simply blind and can't see the wood for the trees.... 😞

[ttokoro]

D, A and E are collinear so the area of Triangle BED is divided by S(B,D,A):S(B,A,E)=7:4.

ABC is mirrored at AB to ABE. So, S(A,B,C) is also 4. Therefore, S(A,C,D) is 7-4=3. 

B, C and D are collinear so the area of Triangle ABD is divided by S(A,B,C):S(A,C,D)=4:3. Therefore, BC:CD=4:3.

Using both area and line length rasio of triangle with same hight is the tips. 

Tokoro.

[Werner_E]

Thanks, using the ratio of the areas additionally is a clever idea. Personally I'd prefer using the intercept theorem as shown above, but this may be a matter of personal taste.

In any case, this is a nice task - thanks for it.