proof-the-circles-in-the-diagram-are-concentric-and-overlinemathrmfg-cong-overlinemathrmgh-prove-tha

Name: proof the circles in the diagram are concentric and overlinemathrmfg cong overlinemathrmgh prove tha
Uploaded: 2019-01-20
Duration: 5 min 6 s
Channel: Numerade
Description: PROOF The circles in the diagram are concentric and $\overline{\mathrm{FG}} \cong \overline{\mathrm{GH}}$ . Prove that $\mathrm{HK}$ and $\mathrm{NG}$ have the same length.

Question

PROOF The circles in the diagram are concentric and $\overline{\mathrm{FG}} \cong \overline{\mathrm{GH}}$ . Prove that $\mathrm{HK}$ and $\mathrm{NG}$ have the same length.

Wen Zheng · Accepted Answer

it&#x27;s a problem. This that kept the r and smaller bee the righty off this segment. Scrubbed and scrubbed circles of Triangle ABC respectively. And let s is equal to and plastic pussy over to at a true that you are going to a over science a zero to be overseeing of b. Is he going to see over saying off? See here a scotch triangle. So this a b c and this is have to r So behalf this angle is there going to too? Eh? And here how this andl it&#x27;s equal to Hey now let&#x27;s look at this small Trang go. So we have her pettiness. Is he going to capital? Are this angle is equal to a thiss part? Is he going to our times? Sign off, eh? Which is he going to? Well, half off the last ofthe bc This is my half off the side, eh? So eh over sigh of a Is he going to to our similarly we can compute What you are is also Iko too Be silent be which is also equal to see over sand Sea So the cows are here because they use lof size I&#x27;d be true that smart is a call to this one. So first I scared you, right? Triangle here. If this is smaller Smarr Smarr, it&#x27;s a B C. No, Catherine, this three Chang goes so half the area off the triangle. ABC Is he coming too? Well, half time. Speed him. Times are last half come speed terms are last one half times. See, terms are which is it? Two, as times are. So we also half area off ABC Is he going to score it? I stamps as minus a. I asked my speed hams as minus C from this too. How iced times are physical, too square root of S Tom&#x27;s past minus k. I have asked minus b ham&#x27;s testaments. See the smaller is equal to divided by us for process. This is a discarded as fine a stay ham&#x27;s cry some honesty hams as mosi over us

Ahyeon Hwang · Answer

number 10 were given circle. Oh, and cord A B is parallel to court CD. You have to prove that arc A C is congruent toe arc, Petey. It&#x27;s the first week in draw straight line through it the center and then draw for Rady. I connecting to all the points this makes my sauce Elise to. I saw Seles trying calls so this side is continuing to the side and the scientist could grew into the side, which means that these angles are the same and these angles are also the same. That means that thes angles are also the same. Therefore, a C arc A C is congruent to Arc B D.

Tony Weisse · Answer

we&#x27;re gonna prove the angles outside the circle Therm noticing each of these three cases. We have the point outside the circle. Here. Here we have two tangents, and here we have to seek ants. So the 1st 1 we had a tangent and a C can&#x27;t. So in each of these three cases, we&#x27;re gonna look at the triangle and angle one is a tangent and see can&#x27;t. And I&#x27;m drawing this corden. There we go. An angle to what we have is the measure of angle to It&#x27;s an extra year angle of this triangle, and that equals the sum of the measure of angle one plus the measure of angle B. Now what we can look at is angle, too. It&#x27;s a tangent cord that Vertex is on the circle, so it&#x27;s half of that arc. So the measure of angle to is 1/2 of our but you see and the measure of angle B over Texas on the circle. So it&#x27;s half of this air measure of Angle B is half of the measure of a C. So let&#x27;s do two things here. First of all, subtraction property, I can get the measure of a single one equals the measure of angle to minus the measure of angle B. So all I did was subtracted be over to this side and then I&#x27;m gonna substitute these in the measure of angle, too. I have is 1/2 of the measure of our BC and the measure of angle B iss 1/2 the measure of R. K. C. So using the distributive property, the 1/2 is common to those. And we have the measure of B C minus the measure of a C angle. One is half of those arcs. So that&#x27;s what we wanted to prove for this first case. The second case. We&#x27;re gonna do the same thing. Looking at this triangle, we have this exterior angle. Let me call that when they call this point. Yes. So we have the measure of angle. P. R. S equals the sum of these two remote interior angles, so it&#x27;s angle too, plus angle P. Now, in this case, I want to solve for angle too. So I know the measure of angle, too. I&#x27;m gonna subtract P so I get the measure of angle P. R s minus the measure of angle P angle. PRS for Texas on the circle. It&#x27;s made by a tangent and accord, so it&#x27;s half of it&#x27;s intercepted our the measure of angle p at work, Texas on a circle. So it&#x27;s half of it&#x27;s intercepted Eric. So the measure of Angle P is equal to half the measure of PR. So substituting in for this equation here I&#x27;ve got the measure of angle, too, equals PRS, which is half of the measure of our peaky wire and minus the measure of Angle P, which is half of the measure of Arc PR. So the measure of angle to distributing out the 1/2. I get 1/2 the measure of our P Q R minus the measure of our PR going to the third case in the same type of thing. I&#x27;m going to start off with this triangle, and we have an extra year angle. So we have the measure of angle X Z. Why it&#x27;s an exterior angle said equals to some of these remote in tears. The measure of angle three plus the measure of angle X. So the measure of angle three would equal. I have to get rid of this ex so I get the measure of angle X z. Why? Minus the measure of Angle X and then, like we did in the other ones, the measure of angle X Z. Why this angle here, the Vertex is on the circle. So it&#x27;s half of this ark and the measure of angle X. The Vertex is on the circle. It&#x27;s half of this ark. So now that we have that, we&#x27;re gonna rewrite this equation. We have the measure of angle three instead of x Z. Why, we&#x27;re gonna put in equal toe half of the measure of X Y minus instead of the measure of angle X, we&#x27;re gonna put half of the measure of W Z. And, like I didn&#x27;t the other ones, there&#x27;s a 1/2 common. When I factor that out. I&#x27;m left with the measure of X y, minus the measure of W. C, which is what we&#x27;re trying to prove

Heather Zimmers · Answer

all right, This problem is a proof, and what I&#x27;m planning to do is give my statements and my reasons. But I&#x27;m not using any particular proof format, so you can translate into a two column proof if that&#x27;s your preference, or you can translate into a paragraph proof if that&#x27;s your preference. So we&#x27;re given a circle, and we have a diameter that&#x27;s perpendicular to accord. And we&#x27;re trying to prove that it separates the cord into two congruent parts and also separates the corresponding arc into two congruent parts. So basically, we&#x27;re trying to prove that the diameter sex, the cord and by sex the angle so we always wanted list are givens first, So we have r E g is the diameter maybe abbreviate a little bit? Uh, circle? No. And we have e g is perpendicular to D F given. All right, now, what does it mean to be perpendicular? It means that they&#x27;re at right angles. So we know that angle D C E o and angle F c l are right angles. Definition of perpendicular is our reason. And if they&#x27;re right angles, that means we have right triangles. So we can say that Triangle D C L and Triangle F C l are right triangles. By the definition of a right triangle, it has a right angle. And the reason I want to state that we have right triangles is because we can prove triangles can grew in by h l. If we&#x27;re using right triangles. So that&#x27;s my goal is my goal is to prove that these two triangles are congruent and so I need to establish some congruent parts. So we know that C L is congruent to itself by the reflexive property so we can list that CEO is congruent to CEO reflexive. And we also know that d l is a radius and f l is a radius and already I in a circle are congruent So deal is congratulated to f l because already I are congruent. Okay, so now we have to right triangles with which have congruent pair of legs and it can grow a pair of high pot nooses so we can claim that the triangles are congruent by high potting use. Leg triangle d c l is congruent to triangle f c l by the high pot news leg congruence of triangles, right triangles Now that we&#x27;ve established that we have congruent triangles, then we know that corresponding parts of them are congruent and the corresponding parts were interested in our segments D c and segment FC So D c is congruent to FC and you could say CD instead of instead of d. C if you want. Um, by corresponding parts, uh, congruent triangles are congruent, which you might know as C p c T. C. It&#x27;s an abbreviation for that, and another one that&#x27;s congruent for the same reason that we might be interested in is angle D l C is congruent to angle f l c by the same reason. Corresponding parts. Okay, so part of what we set out to prove has been finished. The segments are congruent to each other and then we can use the other the angle congruence here, this angle Congrats to this angle to show that this arc is congruent to this ark because we know the measure of the angle is equal to the measure of the ark by definition. So our final statement is that, um, Arc de g is congruence to arc f G because of definition of arc measures. We already established that their central angles are congruent. Therefore, the the ark&#x27;s themselves air congruent. Now perhaps you would break this last part up into a few more steps. Could expand on this idea. Depends on how technical you like to get with your proofs.

Taylor Shimono · Answer

So for this problem, we&#x27;re going to be able to prove the in center throughem using angle congruence. Um, and then also so triangle Congrats. So first, we&#x27;re going to start by looking at, um, some of our angles, um, that deal with Ojai. Let it, um, in green with these two triangles right here. Um triangle a g and a d d. I&#x27;m so since we know that a D is the by sector of angle see a beat that tells us that we&#x27;re going to be able to say that angle C a d is congruent to angle b a d um And so those air those two angles right there, Um and so next we&#x27;re going to be able to say, um that since we know that uh, g g is perpendicular to see a which were given. And we also know on that d e is perpendicular to a B. Uh, we know that these two angles are the same. And also since this side is shared right here and it goes from point A to point D, we know that that&#x27;s the same side. So that tells us that we have to control and angles in one can grow inside. I&#x27;m so bye the angle angle side congruence that is going to allow us to state that triangle e d is congruent to our other triangle triangle a g d. And so by the concurrent triangle through him all right, Not if you&#x27;re in, but just by the fact that they are can grow it. We know that side&#x27;s d E and A d and d g are going to be congruent. Um, so d g in t e we said because of the fact that they are corresponding parts of congruent triangles and so we could go ahead and use this same process with the fact that a B D by sex angle C B A. I blew it again. Say that these two angles are the same. These two angles on the same in this side is the same so again, a bike in growing. But the fact that they&#x27;re congruent triangles and we would be able to find that DF is can grow to d. Um And so since I already said that d e and G G or can grow, we know that this is also going to be congruent to line d G um, which is exactly the proof of the

Problem

REPEATED REASONING $\overline{A B}$ is divided in…

Problem 41 Easy Difficulty

PROOF The circles in the diagram are concentric and $\overline{\mathrm{FG}} \cong \overline{\mathrm{GH}}$ . Prove that $\mathrm{HK}$ and $\mathrm{NG}$ have the same length.

Answer

Therefore, it is proved that $m \widehat{\mathrm{NG}}=m \widehat{\mathrm{JK}}$

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Therefore, it is proved that $m \widehat{\mathrm{NG}}=m \widehat{\mathrm{JK}}$

Geometry A Common Core Curriculum

Catherine R.

Kristen K.

Samuel H.

Michael D.

(A few) Circle Theorems

Angles and Cirlce Theorems with chords and tangents

Ron Larson, Laurie BoswellGeometry A Common Core Curriculum

Catherine R.

Kristen K.

Samuel H.

Michael D.

Ron Larson, Laurie Boswell

Geometry A Common Core Curriculum