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

1. 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 $DG = AB$.

2. anonymous

3. anonymous

all i know is that BF = AE and BCF and ABE are congruent

4. anonymous

i also cannot decide if BG is perpendicular to AE... or not?

5. imqwerty

well this is a pretty simple question if ur using co-ordinate geometry. :) |dw:1440261844607:dw|

6. imqwerty

the equation of a line passing through points A(a,b) and B(m,n) can be written as - $\frac{ y-b }{n-b}=\frac{ x-a }{ m-a }$

7. 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...

8. imqwerty

ok lemme think that way :)

9. imqwerty

we can use trigo right?

10. anonymous

we havent gotten to trigo yet...

11. dan815

12. anonymous

sorry, how do we know that angle BEA is 60 degrees?

13. ganeshie8

scratch that, thats not 60, lets start over

14. ganeshie8

|dw:1440263391261:dw|

15. ganeshie8

so your earlier observation still holds : BG is perpendicular to AE...

16. anonymous

@ganeshie8 where do i go from there? I can make a series of expressions for each angle..but would that really help?

17. 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|

18. anonymous

Thanks so much @ganshie!

19. ganeshie8

np