In the diagram, what is the length of AB?

Hi Mike - In the question that's in the link that you have posted, I am struggling to get how do you get B?

Hi Bunuel - will it possible for you to explain how you make the sides proportion and which sides to make proportionally equal? Is it base to base and height to height ? I hope I am making myself clear. Also, any idea how to solve the question in the link that Mike has provided?

Check this post: length-of-ac-119652.html?hilit=corresponding#p1029262

The ratio of the corresponding sides are equal (corresponding sides are the sides opposite the same angles).

Dear enigma123,

In the problem at the link: https://gmat.magoosh.com/questions/1022,

Because both Triangles ABE & ACD share (1) a right angle, and (2) angle A, we know they are similar.

Because B is the midpoint of AC, we know 2*(AB) = AC, and thus, the ratio of the two similar triangles is 1:2. In other words, the ratio of any two corresponding sides is equal to 2.

CD/BE = 2, and if we know the length of BE, we can solve for the length of CD. Did you submit your answer and watch the video explanation to the question?

Does what I explain here make sense? Please let me know if I can answer anything else.

Mike

The question doesn't even talk of being a right angle triangle. Should we assume the details just because we see it in the image ?

Square signs above (in green circles) indicate that the angles at them are 90 degrees.

Would you assume the lines as parallel in this question ?

Well, the second statement directly gives that DE is parallel to CA (DE||CA), but for the first statement we cannot assume that.

Hey Bunuel,

When you extend a line from the vertex of a right angle to the hypotunse, each of the new two triangles are similar to the larger one. But are the two smaller triangles similar to one another?

Yes, the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Generally if two triangles are similar to the third one, then they are naturally similar to each other.

The key ingredient needed to solve this question is triangle similarity.
Two similar triangles have all the angles congruent. Since the sum of the angles in a triangle is always 180,
knowing that two pairs of angles in the two triangles are congruent is enough to ensure similarity.
To prove that the height in the right triangle ABC determines two similar triangles, look for the congruent angles.

In the given question, even if you don't know the property that the height to the hypotenuse divides the triangle into two similar triangles (and each of them is similar to the initial right triangle), you can still prove that triangle ABD is similar to triangle BDE:
Both are right angled triangles. In addition, because AB is perpendicular to BC and so is DE, AB is parallel to DE, so angle ABD is congruent to angle BDE.
Two pairs of congruent angles means the two triangles are similar.

Knowing all the sides in the triangle BDE, then writing the proportions and because BD belongs to both triangles, you can find AB.

Hey Bunuel,

Thank you very much for your response. I guess then as corrolary to the first question. Since all the angles are the same in all three triangles (let say angles x*,y*, and z*, where z*=90), Is there an easier way to identify which angle is referring to which side in the triangle within the equation you stated, AB/AC=AD/AB=BD/BC? Or are you mentally rotating the triangles making sure angle y* is opposite to AB, angle x* is opposite to AC, and checking it with angle y* opposite to AD, and angle x* is opposite to AB. Are you mentally rotating and checking that the all the sides are in conjunction with (opposite to x*)/(opposite to y*). Or is there an easier, more systematic method your using to ensure that all those sides are in fact in the same proportion.




On another side note, unrelated to my question above:
I also noticed that in your equation that each ratio is of sides of the same triangle equaling the ratio of the other triangle, opposite degree sides. So (Side 1, Triangle A)/(Side 2, Triangle A)=(Side 1 Triangle B)/(Side 2, Triangle B). I always imagined that if Triangle A (sides X,Y,Z) and Triangle B(sides M N P), where triangle A is bigger than B, then their similarity defined the following relationship:

X=Mc , where c is a some positive constant>1 and sides X and M are opposite to the same angle
Y=Nc
Z=Pc

Then the relationship would be defined as c=X/M=Y/N=Z/P= (side 1 Triangle A)/(side 1 Triangle B)=(side 2 Triangle A)/(side 2 Triangle B)=(side 3 Triangle A)/(side 3 Triangle B)
I just noticed that this relationship can always be manipulated to get c=X/M=Y/N ==> new constant=M*c/Y=X/Y=M/N=(Side 1 Triangle A)/(Side 2 Triangle A)=(Side 1 Triangle B)/(Side 2 Triangle B), which i think is very interesting. Same sides of the triangle are also in proportion to the same sides of the other triangle and another new constant the related the two triangles is embedded within the relationship of the triangles own sides!!! (eek, hope that made sense)

I also learned from you that if you had two similar triangles then (Area Triangle A)/(Area Triangle B)=(Side 1 Triangle A)^2/(Side 2 Triangle B)^2.

Are there any other key relationships that are important about similar triangles?

Hi Bunuel,

I understand that a line perpendicular to a hyp will always create TWO similar triangles. That being said, line BD creates ADB and BDC. Correct?

Now, we have another segment DE, since it's perpendicular to BC, it will create BED and DEC. Correct?

Are BED and DEC similar to ADB and BDC or are the two groups standalone? The reason why I ask is because they are both segmented off another hypotenuse?

Angles in ABC are identical to those in ABD and BDC.
Angles in BDC are identical to those in BDE and DEC.

Obviously, angles in all those triangles are identical.

Hi Bunuel,

Can we do this by using isosceles triangle property for triangle ABD in which angle ABD is 45 ( BD is dropping a perpendicular ) so angle A is 45 hence AD is also 5 and we can calculate AB via Pythagoras and same can be done for remaining angles as well.

BD is not a bisector of angle B, so ABD is NOT 45 degrees. BD would be a bisector if AB were equal to BC in this case the perpendicular from B will coincide with the bisector from B.

I have solved this one , but I have one question though... It was not really important (because its a DS) which sides to (short or long leg) to write in tha ratio, BUT for a PS question it's important. HOw one can know which sides are corresponding ? In this picture it's not easy to identify short and long leg....?