HELP!!!!

[potent escarp]

The problem says: In the image ΔABE≅ΔBCD. Calculate the length of x°.

[potent escarp]

I don't know how to start

[west wave]

is it where AEB~DCB, or where AEB~BCD?

[potent escarp]

No no they're congruent no similar

[west wave]

that's what i mean

[west wave]

ABE≅BCD is ambiguous here

[potent escarp]

What do you mean?

Why ambiguious?

[west wave]

Is is that BE=BC and AE=DC, or that BE=DC and AE=BC?

or is it that AE=EB=BC=CD?

[potent escarp]

No no it only says that they're congruent

But it doesn't says which sides

[west wave]

Maybe it doesn't matter?

idk

[potent escarp]

Maybe

Well, how would you start?

Because i don't see nothing, maybe by the congruence, BA=BD, BE=BC and AE=CD

[west wave]

yeah that's how i interpreted it

what have you done?

[potent escarp]

Nothing man

I don't see nothing

[west wave]

hmm

[potent escarp]

Maybe angle BAD=CBD by alternates?

[west wave]

they're not alternate angles tho

[potent escarp]

Oh.

[west wave]

i'd recommend like

labelling the figure with everything you know

[potent escarp]

Well

Maybe angle EFD=BFC by angles oppositve by vertex

Also BFE=CFD by the same

Then nothing man

Are the triangles BFE and CFD congruent?

[coarse grotto]

Are the angles for ADB and DBC the same?

I meant ADB my bad

[potent escarp]

Well the problem doesn't say nothing, but @west wave says that no.

[west wave]

we dont know AD||BC, right?

don't*

[potent escarp]

No, the problem doesn't say nothing.

But it seems that yes

[west wave]

it looks like you're just dealing with angles rn

what do you know about angles in congruent triangles?

[potent escarp]

I think yes

[west wave]

like what other angle is equal to ABE?

[potent escarp]

By congruence BCD

[west wave]

to abe?

[potent escarp]

Oh no no

Sorry

I thought you mean AEB

CBD maybe is equal ABE

[west wave]

yeah

and you can do that two more times

to get two more equal pairs of angles

you might also find it helpful to label which sides are equal too

[potent escarp]

So you are meaning that ABE is 45-45-90?

[west wave]

no

[potent escarp]

Ok

[west wave]

what do you have right now?

as a figure

[potent escarp]

I have like a trapezoid?

[coarse grotto]

Is number 2. supposed to be blank there?

Below the ABE = BCD

[potent escarp]

Yes

[west wave]

i mean

like

if you label everything on your figure that we've just talked about

like sides and angles that are equal

then what do you see?

or like

what does the image look like now?

[west wave]

i'm trying to get you to make something looking more like this

[coarse grotto]

AB = BD, so if we look at triangle ABD it’s a isosceles

[potent escarp]

True

[coarse grotto]

Angle A and D should be the same no?

[potent escarp]

Oh

[coarse grotto]

But we only know part of D

[potent escarp]

Yes and the other what you said yes

[west wave]

angles BAD and BDA have the same measure

[potent escarp]

Yeah by isosceles

[west wave]

you should get something like this right?

[potent escarp]

It won't be all the vertex A 70°, not only BAE?

[coarse grotto]

Same thing with BE= BC. So BEC= BCE

[west wave]

BAD is 70, not BAE

[potent escarp]

Yeah I mean that

[west wave]

but this is interesting

BAD is isoscoles

isoceles*

can you see any other triangle that is isoceles?

that might help us find x?

[coarse grotto]

BEC

He marked that already

[west wave]

right

do you see what to do next?

[potent escarp]

Maybe BFE and CFD are congruent, if then BCD is isosceles no?

[west wave]

that's true

[potent escarp]

or CFD and BCF maybe are congruent

[coarse grotto]

CE isn’t perpendicular tho

I don’t think so

[west wave]

it might be helpful to look back at BEC again

what other triangle might that one be congruent to?

in other words, what triangle do you want BEC to be congruent to?

[potent escarp]

I don't know.

[coarse grotto]

AEB?

[west wave]

why would you want that? /lh

i mean

like

idk how to motivate this step

||if BEC~BAD, then we'd be done, x=70.||

||so it'd be nice for BEC~BAD||

[coarse grotto]

I was looking at that

Yes

[potent escarp]

They'd be similar.

But not congruent.

[west wave]

yeah

if BEC and BAD were similar, then we're done

now all that remains is to prove it

[coarse grotto]

Angle CBE= ABD

[west wave]

yup

why?

[coarse grotto]

ABE and FBC are equal. ABE+ EBF= ABD likewise for the other one

[west wave]

right

[potent escarp]

yes

[west wave]

then you just have to prove it

[coarse grotto]

My proofing language might not be correct because I haven’t taken geometry yet so I am trying

[west wave]

like

[west wave]

ah

there's two ways i guess to think about it

you can use BAD being iso. to find ABD, say it's EBC, then use triangle EBC being iso. to find x

or you can just say "the two triangles are similar"

i don't recall if the second one is valid to say

but it's true

[coarse grotto]

Ok

[coarse grotto]

Do you understand?

[potent escarp]

Yes yes

The problem is just solved?

Or are you still thinking?

[west wave]

well

i mean

the problem is solved, in the sense that the proof has been broken down

what we've said is not to the level of rigor or technicality that you'd need to write

[potent escarp]

Ohhhhh

That's why I was confused

[west wave]

alr

[potent escarp]

But wait wait

[west wave]

yeah?

[potent escarp]

I didn't understand the part of similarity

[west wave]

ah

how BAD is similar to BEC?

[potent escarp]

AA?

[west wave]

on a higher level, it's kind of SAS similarity

oh wait

[potent escarp]

well that's not a criterie of congruence?

[west wave]

that's a thing

[potent escarp]

What?

[west wave]

if it were true that BA=BE and BD=BC, then we'd want to use SAS coongruence

congruence*

but all we have is BA/BD = BE/BC

that's good enough for SAS similatiry

similarity*

[potent escarp]

So you're meaning that the other traingle also is rectangle?

is a right triangle?

[west wave]

wdym?

we're taking about BAD and BEC

being similar

[potent escarp]

Yeah yeah

Sorrry

I saw bad

So I have all i think can you please wait me. i'm going to write it in the correct language.

[west wave]

alr

[potent escarp]

Please gimme 15 min.

[west wave]

lmk when you're done

[potent escarp]

K

[potent escarp]

@west wave sorry but i only that the angle CEB and ABD are equal. I don't see the other sides how they're similar. well i have never prove similarity in sides how i do that?

[west wave]

you could show that BA/BE=BD/BC

alternatively, you could solve for the angles in both triangles and use AAA...

idk

[potent escarp]

So if i want to do with sas then i only put BA/BE=BD/BC?

[west wave]

yeah

[potent escarp]

Only that?

Without proof?

[west wave]

if you show BA/BE=BD/BC and CEB=ABD then you can cite SAS

[potent escarp]

So you mean CBE=ABD?

Or as isosceles 30°=30° ?

[west wave]

right yeah

[potent escarp]

But then BA/BE=BD/BC, maybe if i show parallelism?

[coarse grotto]

SAS could work

[potent escarp]

Yes, but how do i prove that BA/BE=BD/BC?

[coarse grotto]

I am gonna repost this because it’s so far up

[potent escarp]

I have this so far

[west wave]

i agree

two things

the "the" at the start of your lines is unnecessary

and you dont need line 2

[coarse grotto]

True

[potent escarp]

Ok

[coarse grotto]

I learnt more geometry here than I learnt in my lifetime so far

[potent escarp]

LOL

[coarse grotto]

I am taking geometry later this year though

So this experience will definitely help

[potent escarp]

Yes, are you in 7th grade?

[coarse grotto]

No I am actually in high school. I just never did geometry and skipped to algebra 2 and pre calc

I am doing pre calc first semester and geo second

[potent escarp]

Oh I see

But in statement 4 i don't what to put besides the angles, how to prove both are similar?

[coarse grotto]

Say therefore angles ADB= BCE

[potent escarp]

I just wrote it

[coarse grotto]

Because the triangles are same

[potent escarp]

So it'll be congruence no similarity?

[coarse grotto]

No you said the top angles are the same

I said 70 is x

[potent escarp]

True but that's for the final

i need to prove how BA/BE=BD/BC

People pls

[short yarrow]

BF/BC* ?

is that what you meant

[potent escarp]

Bro to be hoenst

They told me that

[short yarrow]

CBE = DBE + CBD = DBE + ABE = ABD

[potent escarp]

True

[short yarrow]

?? what is that second claim

that's true but it doesnt prove similarity?

[coarse grotto]

Didn’t you already really prove BA/BE=BD/BC in line one or am I wrong

[short yarrow]

that's obvious from the data

they're congruent triangles

[coarse grotto]

Then why did you state it agian in line 4

[potent escarp]

Man, I'm confused are they congruent or similar?????????

[short yarrow]

what you want to prove is that BA/BD = BE/BC

[potent escarp]

YESSSSS

[west wave]

he's saying this fact is obvious because AEB is congruent to DCB

[short yarrow]

how does it help though

[west wave]

but it is used to prove that BAD and BEC are similar

[short yarrow]

which is also obviously true because of congruent triangles

BA/BD = hyp/hyp
BE/BC = side/side

and since they're both congruent triangles it's true

so the triangles are similar and BCF=70

@potent escarp did you understand

[potent escarp]

So,so

I have this

[short yarrow]

B, E F are collinear by construction no?

[west wave]

sup

[short yarrow]

Is f not defined as CE intersection BD?

[west wave]

yup*

[short yarrow]

yeah so BCF = BCE = xº = 70º

that's it

[potent escarp]

But my reasoning is good right?

[coarse grotto]

Don’t try to overthink

[potent escarp]

Oh ok.

[coarse grotto]

Once you have the answer reasoning is pretty straightforward

At least to me

[potent escarp]

Oh ok.

Thanks @west wave @short yarrow @coarse grotto

Thank you very much

https://tenor.com/view/the-office-michael-scott-thank-you-bow-steve-carell-gif-5009516783019270794

.solved

[neon basalt]

Solved

Post marked as solved by @potent escarp.

Use .unsolved if this was a mistake.