Lesson: Similar Triangles

Comment on Similar Triangles

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
gmat-admin's picture

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


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

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
gmat-admin's picture

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


Your solution is so simple and amazing. Thanks!

sir can we assume triangle to be equilateral if nothing is given?
gmat-admin's picture

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!
gmat-admin's picture

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

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!

gmat-admin's picture

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?


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.
gmat-admin's picture

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,


This is the first Q in the reinforcemen t activities .inks list. Is there a way to solve it with similar triangles?
gmat-admin's picture

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 ?
gmat-admin's picture

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


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?
gmat-admin's picture

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?

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