HELP!
Given;

Question

HELP! 
Given; <ONL is congruent to <MLN , <O and <M are right angles.
Prove; LM is congruent to NO.

anonymous · Answer

attachment

anonymous · Answer

Given. All right angles. LN is congruent to LN. -> ? ONL is congruent to MLN -> ? LM is congruent to NO -> ?

uri · Answer

I suck at Geometry,Sorry.

sleepyjess · Answer

sorry i am horrible at proofs

anonymous · Answer

ok. thanks anyway!

mathmath333 · Answer

in $\triangle LNO $ and $\triangle NLM $
$\large\tt \begin{align} \color{black}{\angle LNO \cong \angle MLN------(given)\~\
\angle LON \cong \angle LMN------(90^{\circ})\~\
side LN \cong side LN------(same ~side)\~\~\~\


}\end{align}$
therefore
$\triangle LNO \cong \triangle NLM $

by $\Large SAA $ theorm


hence we can say that 
$\Large side ~NO \cong  side ~LM$

anonymous · Answer

thanks sooooooo much, can help with another problem please? @mathmath333

mathmath333 · Answer

i mean bring it on lol

uri · Answer

@mathmath333 are you an engineering student? :p

mathmath333 · Answer

yes how do u know that

anonymous · Answer

Given: ABCE is a rectangle. D is the midpoint of CE. Prove: AD is congruent to BD 1. ABCE is a rectangle. D is the midpoint of CE. -> Given 2. Definition of rectangle 3. AE is congruent to BC -> Definition of rectangle 4. ? -> ? 5. ? -> ? 6. ? -> ?

anonymous · Answer

attachment

uri · Answer

only they understand this. Mostly. :p

mathmath333 · Answer

this stuff a used to prove when i was in 8 th 9th grade ,
engineering applied math is way tough than this

uri · Answer

you're in which grade now?

uri · Answer

I was a commerce student so I had business mathematics,statistics,economics and accounting.

uri · Answer

Now i'm becoming a designer...xD

anonymous · Answer

the second problem is posted.  @mathmath333

mathmath333 · Answer

in $\triangle AED $ and $\triangle BCD $
$\large\tt \begin{align} \color{black}{\angle AED \cong \angle BCD------(90^{\circ}~given~:\rm \text{defn of rectangle})\~\
side AE \cong side BC------\
---(\text{opposite sides of rectamgle are congruent})\~\
side ED \cong side DC------(\text{as D is the midpoint of EC})\~\~\~\


}\end{align}$
therefore
$\triangle AED \cong \triangle BCD $

by $\Large SAS $ theorm


hence we can say that 
$\Large side ~AD \cong  side ~BD$


i already graduated as an engineer few months ago
nice  designing :)

anonymous · Answer

thanks so much! @mathmath333