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Mathematics 13 Online
simsharrison:

Geometry?

simsharrison:

simsharrison:

jimthompson5910:

hint: XYZ is the shared overlapping angle between the two triangles

simsharrison:

this is what I put in: The congruence theorem would be Side-Angle-Side proving triangle XBY is congruent to triangle ZAY.

simsharrison:

for 2b.

jimthompson5910:

We're already given \( \angle X \cong \angle Z\) and \( \overline{XY} \cong \overline{ZY}\) That takes care of the "A" and "S" respectively As mentioned with the overlapping angles, if we knew that \(\angle XYB \cong \angle ZYA\), then we would have enough information to use ASA.

jimthompson5910:

Once we've proven \(\triangle XBY \cong \triangle ZAY\), we use CPCTC to lead to \(\overline{AZ} \cong \overline{BX}\) CPCTC = corresponding parts of congruent triangles are congruent

simsharrison:

oh its ASA and not SAS?

jimthompson5910:

correct

jimthompson5910:

|dw:1607649552118:dw|

jimthompson5910:

We know angle X is congruent to angle Z |dw:1607649621800:dw| So that's one "A" of ASA

jimthompson5910:

and we know that XY and YZ are the same length That's "S" of ASA |dw:1607649695012:dw|

jimthompson5910:

If we knew that angle XYB = angle ZYA then we have enough for ASA |dw:1607649744356:dw|

jimthompson5910:

note how the marked sides (the sides with the double tickmarks) are between the marked angles |dw:1607649823980:dw| So that tells us we aren't using AAS but instead we use ASA

simsharrison:

ok thank you @jimthompson5910

jimthompson5910:

No problem

simsharrison:

do you have time for 2 more? lol if not that ok.

jimthompson5910:

sure

simsharrison:

simsharrison:

jimthompson5910:

how far did you get with problem 1?

simsharrison:

I have only done the given's, I haven't got farther than that I am stuck

jimthompson5910:

What do you notice about ZX?

simsharrison:

its in the middle but also a bisector. Is that what you mean?

jimthompson5910:

it might help to break up the triangles like so |dw:1607650121667:dw|

jimthompson5910:

Would you agree that ZX = ZX?

simsharrison:

yes by reflexive prpperty

simsharrison:

property*

jimthompson5910:

yes

jimthompson5910:

|dw:1607650229356:dw|

jimthompson5910:

Do we have enough to prove these triangles congruent?

simsharrison:

yes by SAS

jimthompson5910:

yep

simsharrison:

But that was in the given already that those two trangles are congruent

jimthompson5910:

Then by CPCTC, we know that angle WXZ = angle YXZ |dw:1607650308530:dw|

jimthompson5910:

but those angles I just marked also add up to 180 since angle WXY of the original drawing is a straight angle let p = measure of angle WXZ = measure of angle YXZ You need to solve p+p = 180

simsharrison:

by triangle sum theorem

jimthompson5910:

what do you get when you solve for p in the equation p+p = 180

simsharrison:

ill be honest idk im lost...

jimthompson5910:

p+p turns into 2p, correct?

simsharrison:

oh.. yes sorry

jimthompson5910:

so if 2p = 180, then p = ??

simsharrison:

i am this far so far

simsharrison:

p = 90

simsharrison:

oh so right angle

jimthompson5910:

that takes care of the "perpendicular" part

jimthompson5910:

to prove the "bisector" part, you need to show that WX = XY

jimthompson5910:

since "bisect" means "cut in half" or "cut into two smaller equal pieces"

simsharrison:

yes, sorry I went afk but im back

jimthompson5910:

its ok

jimthompson5910:

so how can you show that WX = XY ?

simsharrison:

transitive?

jimthompson5910:

go back to this drawing |dw:1607655359247:dw|

jimthompson5910:

We've proven the triangles to be congruent, so we know that the corresponding pieces WX and XY are the same length |dw:1607655379463:dw|

jimthompson5910:

Because WX = XY, this means segment ZX bisects (cuts in half) segment WY

simsharrison:

oh by the HL theorem

jimthompson5910:

no we proved the triangles congruent by SAS

simsharrison:

yes but isnt the last proof HL

simsharrison:

the bisector one

jimthompson5910:

you only use HL if you know the triangles are right triangles But we already proved them to be congruent using SAS. So using HL is unnecessary

simsharrison:

ok thank you

simsharrison:

that is all I needed, I went ahead and did the other one, thank you so much for your help

jimthompson5910:

You're welcome

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