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## Comment on

Similar Triangles## Hello Brent . One quick

## 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.

GMAT Club: https://gmatclub.com/forum/gmac-official-guides-the-master-directory-lin...

BTG: https://www.beatthegmat.com/official-guide-question-library-t296759.html

Cheers,

Brent

## Oh Brilliant:) That helps .

## 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

## sir can we assume triangle to

## 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, follow up questions

1) how can we assume the triangle to be equilateral when no information is given?

2) how do we know that the the intercepts at points D and E creates a right triangle?

Thank you!

## 1) Since the question doesn't

1) Since the question doesn't mention anything about the kind of triangle we have (isosceles, scalene, etc), then it must be the case that we'll get the same answer for ANY kind of triangle. So, I chose to use a nice, easy-to-work-with shape (equilateral triangle).

2) This is a property of isosceles and equilateral triangles that's covered at 7:05 in the following video: https://www.gmatprepnow.com/module/gmat-geometry/video/865

## 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!

## Question link: https:/

Question link: https://gmatclub.com/forum/in-the-figure-above-a-is-the-center-of-the-ci...

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

## 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,

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?

## You bet.

You bet.

Here's my solution: https://gmatclub.com/forum/what-is-x-in-the-diagram-below-201045.html#p2...

## Hi Brent you mentioned that u

## 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

## Hi Brent,

Can you help me understand how you've concluded that triangle PQS & triangle QSR are similar?

They have a 90 degree angle in common, however, what's the other common angle?

## I believe you're referring to

I believe you're referring to the question at: https://gmatclub.com/forum/in-the-diagram-above-pqr-is-a-right-angle-and...

First notice that triangle PQR consists of three angles, denoted by a circle, star and square (where the square represents the 90° angle)

Triangle PQS consists of an angle denoted by a circle and a square.

This means the third angle in triangle PQS must be a star.

Likewise, triangle QSR consists of an angle denoted by a star and a square.

This means the third angle in triangle PQS must be a circle.

Since triangles PQS and QSR share the same three angles (denoted by a circle, star, and square), we can conclude that those triangles are similar.

Does that help?