Think about triangles...

You have to prove that BE:EX =3:1

Note by A Former Brilliant Member
4 years, 4 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Just construct DF parallel to BX . So by MPT in BXC. DF = 1/2 BX. Now DF // EX. So by MPT in triangle ADF. We get XE = 1/2 DF. i.e. XE = 1/4 BX and so BE:EX = 3:1

akash deep - 4 years, 4 months ago

Log in to reply

Nice solution :).

Karthik Venkata - 4 years, 4 months ago

Log in to reply

You have Given that BD=CD and AE=DE.

A Former Brilliant Member - 4 years, 4 months ago

Log in to reply

If you apply Menelaus' theorem twice you will get that ratio.

Maria Kozlowska - 4 years, 4 months ago

Log in to reply

can we do it by applying mid point theorem?

A Former Brilliant Member - 4 years, 4 months ago

Log in to reply

I do not see how it could be helpful in this case.

Maria Kozlowska - 4 years, 4 months ago

Log in to reply

@Maria Kozlowska draw a line parallel to BC through the point E. say PQ,then in triangle ABD, E is mid point of AD and also,EP parallel to BD,therefore AP=BP,and similarly AQ=QC.Now Triangle BXC is similar to Triangle EXQ, Therefore,BX/EX=XC/XQ=BC/EQ.Let us take,BX/EX=BC/EQ,i.e BX/EX=BC/1/2DC, i.e BX/EX=BC/1/2(1/2B/C),That gives,BX/EX=4/1, Now subtracting 1 from both sides,that gives, BE:EX=3:1

I think Actually we can prove this by 3 ways.

A Former Brilliant Member - 4 years, 4 months ago

Log in to reply

@A Former Brilliant Member I am glad you have found your answers. As always in math, there are many ways to do things.

Maria Kozlowska - 4 years, 4 months ago

Log in to reply

@Maria Kozlowska Yeah thank you.

A Former Brilliant Member - 4 years, 4 months ago

Log in to reply

Karthik Venkata - 4 years, 4 months ago

Log in to reply

yeah but we can also do it by the use of mid point theorem , just draw a line parallel to BC through point E.and further you can solve

A Former Brilliant Member - 4 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...