# Lesson: Similar Triangles

## Comment on Similar Triangles

### Hello Brent . One quick

Hello Brent . One quick questions wrt to the questions you have marked from the OG's for Practice is there somehow I can find solutions to those questions . For example I am using the 15th OG and for questions which I am unable to solve I do not quite understand the explanations given in OG. Please help ### Good question, mimajit.

Good question, mimajit.

GMAT Club and Beat The GMAT have created links to all OG question, and the questions are organized by the edition/year of each particular OG.

Cheers,
Brent

### Oh Brilliant:) That helps .

Oh Brilliant:) That helps . Thanks a ton again.

### Hi Brent,

Hi Brent,

This one is super tough, need your help: https://gmatclub.com/forum/given-a-triangle-abc-where-g-is-the-intersection-of-point-of-medians-286520.html ### TRICKY!!!!!!!!!!!

TRICKY!!!!!!!!!!!
Here's my solution: https://gmatclub.com/forum/given-a-triangle-abc-where-g-is-the-intersect...

Cheers,
Brent

### Your solution is so simple

Your solution is so simple and amazing. Thanks!

### sir can we assume triangle to

sir can we assume triangle to be equilateral if nothing is given? ### If we aren't given any

If we aren't given any information about a triangle's angles or sides, then we cannot assume that the triangle is equilateral.

### Hi Brent,

Hi Brent,

In the question
https://gmatclub.com/forum/in-the-figure-above-a-is-the-center-of-the-circle-df-is-5-and-ef-is-235664.html

I was able to understand the entire solution except for the last bit, the similar triangles. I understood the similar angle property, but here's what I did -
Ratios of the corresponding sides are equal, which means EF/EC=EA/EB => 25/25+x=20/40 from which the value of x came out to be 25. This is based on the explanation that if there are two triangles with sides a,b,c and second triangle d,e,f, then the ratios will be a/d=b/e=c/f
But in the explanation given by you, you used EF/EB=EA/EC. How did you arrive at this? Please explain. They don't appear to be the corresponding sides.
Thanks so much for all the help! To find corresponding sides, we must pay close attention the the angles in each triangle.

In the RED triangle, EF is between the star and the smiley face.
In the BLUE triangle, EB is between the star and the smiley face.
So, EF and EB are corresponding sides

In the RED triangle, EA is between the star and the right angle.
In the BLUE triangle, EC is between the star and the right angle.
So, EA and EC are corresponding sides

So, we can write: EF/EB = EA/EC

Does that help?

Cheers,
Brent

### Hi, Brent. I think that your

Hi, Brent. I think that your comment on making sure that the small triangle must always be on top is inaccurate. I watched study pug and a few other youtube videos on similar triangles and it says that I must maintain the ratios but I can put the small or large on top or bottom as long as I maintain the ratios. ### I don't think I said that the

I don't think I said that the ratios MUST be arranged that format. The only "rule" is that each ratio has the same formal (i.e., large/small or small/large)

As long as the ratios all have the same format, you'll arrive at the correct answer.

### Hey Brent,

Hey Brent,

https://gmatclub.com/forum/what-is-x-in-the-diagram-below-201045.html

This is the first Q in the reinforcemen t activities .inks list. Is there a way to solve it with similar triangles? ### Hi Brent you mentioned that u

Hi Brent you mentioned that u have kept the blue triangle on top and the pink triangle in the bottom to maintain consistency ! Does it mean in the ratio the numerator will always have the sides of the smaller triangle and the big triangle will be the denominator or any order is OK as long as we maintain consitany of keeping the sides of the same triangle in either the numerator or denminator ? ### Either way is fine. All that

Either way is fine. All that matters is that the two ratios must be consistent.
For example, let's say we have a big triangle (with sides A, B and C) and a small triangle (with CORRESPONDING sides a, b and c).

We can say: A/a = B/b = C/c
Or we can say: a/A = b/B = c/C

Cheers,
Brent