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anonymous
 one year ago
last question of the night... promise
anonymous
 one year ago
last question of the night... promise

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In 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
 one year ago
Best ResponseYou've already chosen the best response.0all i know is that BF = AE and BCF and ABE are congruent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i also cannot decide if BG is perpendicular to AE... or not?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1well this is a pretty simple question if ur using coordinate geometry. :) dw:1440261844607:dw

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1the equation of a line passing through points A(a,b) and B(m,n) can be written as  \[\frac{ yb }{nb}=\frac{ xa }{ ma }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is there a way to solve it using properties of quadrilaterals and triangles, because thats what we're working on in class right now...

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1ok lemme think that way :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1we can use trigo right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we havent gotten to trigo yet...

dan815
 one year ago
Best ResponseYou've already chosen the best response.04 angles of a quad add to 360

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry, how do we know that angle BEA is 60 degrees?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3scratch that, thats not 60, lets start over

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1440263391261:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3so your earlier observation still holds : `BG is perpendicular to AE... `

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 where do i go from there? I can make a series of expressions for each angle..but would that really help?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Here 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
 one year ago
Best ResponseYou've already chosen the best response.0Thanks so much @ganshie!
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