Waste less time on Facebook — follow Brilliant.
×

My best 8th Grade geometry problems!

Feel free to post more!

  1. A square \( \displaystyle ABCD \) has \( \displaystyle E, F, G \) are the midpoints of \( \displaystyle AD, DC, AC \). \( \displaystyle GF \) intersects \( \displaystyle AC \) at \(H\). Prove \( \displaystyle EH \perp BH \).

  2. A square \( \displaystyle ABCD \) has \( \displaystyle E, F, G, H \) lies on \( \displaystyle AB, BC, CD, AD \) so that \( \displaystyle AE=BF=CG=HD \). Prove \( \displaystyle EG \perp HF \).

  3. A rectangle \( \displaystyle ABCD \) has \( \displaystyle BH \perp AC \). \( \displaystyle K, E \) are the midpoints of \( \displaystyle AH, CD \). Prove \( \displaystyle BK \perp KE \).

  4. Acute \( \displaystyle \triangle ABC \) with orthocenter \( \displaystyle H \), \( \displaystyle M, N \) lies on \( \displaystyle BH, CH \) so that \( \displaystyle \angle AMC=\angle ANB=90° \). Prove that \( \displaystyle AM=AN \).

Note by Adam Phúc Nguyễn
2 years ago

No vote yet
1 vote

  Easy Math Editor

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 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

2) \(\displaystyle\triangle AEH ,\displaystyle\triangle BFE ,\displaystyle\triangle CGF ,\displaystyle\triangle DHG \) are all congruent.Hence, \(EF=FG=GH=HE\) which implies \(EFGH\) is a rhombus whose diagonals intersect at right angles.

Siddharth Singh - 2 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...