anonymous
  • anonymous
last question of the night... promise
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
In square $ABCD$, $E$ is the midpoint of $\overline{BC}$, and $F$ is the midpoint of $\overline{CD}$. Let $G$ be the intersection of $\overline{AE}$ and $\overline{BF}$. Prove that $DG = AB$.
anonymous
  • anonymous
anonymous
  • anonymous
all i know is that BF = AE and BCF and ABE are congruent

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
i also cannot decide if BG is perpendicular to AE... or not?
imqwerty
  • imqwerty
well this is a pretty simple question if ur using co-ordinate geometry. :) |dw:1440261844607:dw|
imqwerty
  • imqwerty
the equation of a line passing through points A(a,b) and B(m,n) can be written as - \[\frac{ y-b }{n-b}=\frac{ x-a }{ m-a }\]
anonymous
  • anonymous
is there a way to solve it using properties of quadrilaterals and triangles, because thats what we're working on in class right now...
imqwerty
  • imqwerty
ok lemme think that way :)
imqwerty
  • imqwerty
we can use trigo right?
anonymous
  • anonymous
we havent gotten to trigo yet...
dan815
  • dan815
4 angles of a quad add to 360
anonymous
  • anonymous
sorry, how do we know that angle BEA is 60 degrees?
ganeshie8
  • ganeshie8
scratch that, thats not 60, lets start over
ganeshie8
  • ganeshie8
|dw:1440263391261:dw|
ganeshie8
  • ganeshie8
so your earlier observation still holds : `BG is perpendicular to AE... `
anonymous
  • anonymous
@ganeshie8 where do i go from there? I can make a series of expressions for each angle..but would that really help?
ganeshie8
  • ganeshie8
Here is the complete proof : 1) From the below diagram, clearly, \(DH\perp AG\) and \(BG\perp AE\). 2) \(\triangle ABG \sim \triangle AOH\) by \(AA\) similarity, so \(\dfrac{AO}{OG}=\dfrac{AH}{HB}\implies AO=OG\) 3) \(\triangle ODG\cong \triangle ODA\) by \(SAS\) congruence 4) \(DG\cong DA\) by CPCTC. |dw:1440268645673:dw|
anonymous
  • anonymous
Thanks so much @ganshie!
ganeshie8
  • ganeshie8
np

Looking for something else?

Not the answer you are looking for? Search for more explanations.