First one. Find DM + DN in triangle ABC if AB=BC=10 and area of ABC is 50.
I drew the height BK, we know that AB is 10. lets say height is "h" and AK is x. We know that in isosceles triangle height is same as median and bisector. So AK = KC. area of ABC is 1/2 * 2x * h = 50 from here xh=50
also we know that hˆ2 + xˆ2 = 10. But there is no solution. What am i doing wrong
#Several geometry questions.
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Hmm
I don't quite understand your method
Ah, I see.
$h^2+x^2=10^2$, not $h^2+x^2=10$
Kocher
Basically, you forgot to square the 10 when doing the pythagorean theorem
Then this leads you to a singular solution
oh my god bro im so sorry. my brain is not braining
im tired of math
It's fine
I've made worse mistakes 
but i started with right method yeah?
Yes
Spoiler: ||ABC has a right angle at B||
Ok one more thing
yeah because x and h are equal so angles are 45 and 45. and BK is bisector of angle B therefore 45 * 2 = 90
Since there is no given information on MD and DN, we can assume WLOG (without loss of generality) that D is colinear to B
Yes
and D is 90 too
Unless there is given information which you have not provided?
Yes
Hence this
bro i have no idea what to do
@worthy nebula you here
Try what I said earlier
^ (I meant to say that D is the foot of the perpendicular bisector of AC)
i did not get it im sorry bro
wdym
Uh
i really got no idea
i know but i cant. Next week i got an exam. Not school exam but government exam. So not passing it wont get me into any uni. Thats why
yeah i got it man
@worthy nebula
from here i get a = 5. and MD = DN. so 2 * 5 = 10
+close
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