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Mathematics 10 Online
OpenStudy (anonymous):

The figure below shows a circle with center O. Segment PQ is tangent to the circle at P and segment RQ is a tangent to the circle at R.

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

Which statement and reason is most appropriate for box 3?

OpenStudy (anonymous):

Segment OP is congruent to segment OR since they are congruent chords. Angle OPQ is congruent to angle ORQ since they are corresponding sides of congruent triangles. Segment OQ is congruent to segment OQ by the reflexive property of line segments. Angle OPQ is congruent to angle ORQ since they are vertical angles.

OpenStudy (anonymous):

@jim_thompson5910 May you please help me?

jimthompson5910 (jim_thompson5910):

Segment OP is congruent to segment OR since they are congruent chords is false because they are radii (not chords) they are congruent though

OpenStudy (compassionate):

@jim_thompson5910 , any reason I'm blocked?

jimthompson5910 (jim_thompson5910):

Angle OPQ is congruent to angle ORQ since they are corresponding sides of congruent triangles. that's partially true as well: they are congruent, but NOT for this reason (besides this was already proven)

OpenStudy (anonymous):

I was thinking that it was the third choice

OpenStudy (anonymous):

However I am not very good at proofs and I am not entirely sure

jimthompson5910 (jim_thompson5910):

you are correct A$apGeometryQuestions that's a true statement and it's useful in this case

OpenStudy (anonymous):

Thank you so much for validating for me! @jim_thompson5910

OpenStudy (anonymous):

You get a medal.

jimthompson5910 (jim_thompson5910):

and Compassionate, I'm not sure how you got blocked I may have blocked you a long time ago (don't remember doing so though) I'll unblock you

jimthompson5910 (jim_thompson5910):

yw

OpenStudy (anonymous):

May you aid me with one last question please? @jim_thompson5910

jimthompson5910 (jim_thompson5910):

ok

OpenStudy (anonymous):

Give me some time to post it please. Maybe a minute.

OpenStudy (anonymous):

Look at the figure shown below

OpenStudy (anonymous):

OpenStudy (anonymous):

A student made the table below to show the steps to prove that DC is equal to EC.

jimthompson5910 (jim_thompson5910):

post a screenshot of the table

OpenStudy (anonymous):

OpenStudy (anonymous):

Here it is

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

ok notice how lines 4 and 6 have the same right side (after the equals sign)

jimthompson5910 (jim_thompson5910):

what does that mean? what does that allow us to say?

OpenStudy (anonymous):

They are supplementary angles?

jimthompson5910 (jim_thompson5910):

no, nowhere does it say they add to 180

OpenStudy (anonymous):

well the picture slightly suggests it, but I have no idea

jimthompson5910 (jim_thompson5910):

this is a very counter-intuitive idea...but NEVER base any decisions solely on the drawing alone

jimthompson5910 (jim_thompson5910):

the drawing may not be to scale

OpenStudy (anonymous):

Yes, I understand; My fault.

jimthompson5910 (jim_thompson5910):

no worries

OpenStudy (anonymous):

they are corresponding? no?

OpenStudy (anonymous):

ACD and DCE

jimthompson5910 (jim_thompson5910):

ACE and BCD are equal to the same thing, so they must be ______

OpenStudy (anonymous):

congruent/vertical?

jimthompson5910 (jim_thompson5910):

let x = m<ACD + m<DCE

OpenStudy (anonymous):

ok

jimthompson5910 (jim_thompson5910):

so we can say that m<ACE = x and m<BCD = x

jimthompson5910 (jim_thompson5910):

I'm just going to lines 4 and 6 and replacing m<ACD + m<DCE with x

OpenStudy (anonymous):

This is true

OpenStudy (anonymous):

m<ACD + m<DCE=x

jimthompson5910 (jim_thompson5910):

yep

OpenStudy (anonymous):

got it. Next step?

jimthompson5910 (jim_thompson5910):

so if m<ACE = x and m<BCD = x, then what's their relationship?

OpenStudy (anonymous):

they are congruent angles

jimthompson5910 (jim_thompson5910):

yep

jimthompson5910 (jim_thompson5910):

so their angle measures are equal

OpenStudy (anonymous):

this is because both m∡ACD + m∡DCE equals them

OpenStudy (anonymous):

and yes

jimthompson5910 (jim_thompson5910):

so m<ACE = m<BCD

jimthompson5910 (jim_thompson5910):

the reason is ______

OpenStudy (anonymous):

their angle measures are equal

OpenStudy (anonymous):

I'm not sure which postulate I would use to prove it

jimthompson5910 (jim_thompson5910):

well if x = y and x = z, then y = z what property are we using?

OpenStudy (anonymous):

addition property of equality

jimthompson5910 (jim_thompson5910):

no

OpenStudy (anonymous):

Transitive property of equality

OpenStudy (anonymous):

Im almost sure @jim_thompson5910

jimthompson5910 (jim_thompson5910):

closer, but no

OpenStudy (anonymous):

Last option: substitution property of equality, but I doubt it. :S

jimthompson5910 (jim_thompson5910):

well I would just say substitution

jimthompson5910 (jim_thompson5910):

the transitive property is the idea that if x = y and y = z, then x = z so it's very similar

OpenStudy (anonymous):

Ok.

jimthompson5910 (jim_thompson5910):

substitution is the same idea basically

OpenStudy (anonymous):

Yes, it is formally suggested as substitution property of equality in my class,

jimthompson5910 (jim_thompson5910):

I don't think there's a "substitution property of equality" though it's just known as "substitution"

jimthompson5910 (jim_thompson5910):

hmm I guess "substitution property of equality" is a bit valid since it deals with equations

OpenStudy (anonymous):

Yep, It doesn't really make a difference though. It's ok, either way.

jimthompson5910 (jim_thompson5910):

yeah you could argue either way really

OpenStudy (anonymous):

exactly. So would that be the justification used for the answer? "substitution"

jimthompson5910 (jim_thompson5910):

yeah that's what I would ultimately go with: it's short and simple

OpenStudy (anonymous):

So the final answer is: m<ACE = m<BCD ----> substitution

OpenStudy (anonymous):

?

jimthompson5910 (jim_thompson5910):

yep

OpenStudy (anonymous):

I dont get why the proof would not work without that step

jimthompson5910 (jim_thompson5910):

if I were grading this, I'd say it's correct then again, the teacher may be looking for "substitution property of equality" or "Transitive property of equality" but you could argue for either one really

jimthompson5910 (jim_thompson5910):

well look at the next line

jimthompson5910 (jim_thompson5910):

what is that line

OpenStudy (anonymous):

The ASA pos. is saying that the two triangles are congruent

jimthompson5910 (jim_thompson5910):

ok what does ASA stand for

OpenStudy (anonymous):

Angle-side angle

jimthompson5910 (jim_thompson5910):

ok so you need a pair of angles congruent and you need a pair of sides congruent the sides are between the angles

jimthompson5910 (jim_thompson5910):

line 1 gives you the sides line 2 gives you one pair of the angles but which line gives you the other pair of angles?

OpenStudy (anonymous):

the line we just solved

OpenStudy (anonymous):

line 3 also does too i think

OpenStudy (anonymous):

jimthompson5910 (jim_thompson5910):

line 3 does not

jimthompson5910 (jim_thompson5910):

if it did, why did we do all that work after line 3?

jimthompson5910 (jim_thompson5910):

line 6 does though, that's the missing piece of info to complete and set up ASA

OpenStudy (anonymous):

Oh snap lol. I'm foolish. So we then would have all angle measures

OpenStudy (anonymous):

but what about line 7, AKA: the missing line

jimthompson5910 (jim_thompson5910):

oh right, that's what I meant to say

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