circle geometry year 11 extension 1 questions please help! (2 Viewers)

abdullion · Apr 7, 2015

So I've been struggling to answer these two questions, please try help if you can!

1. CD is a diameter of the circle. CE, OF and DG are each perpendicular to the line EXFYG.(X and Y are points on the circle, and EG cuts the circle at two points.) Prove that EX = YG.

2. PAB is an isosceles triangle inscribes in a circle of radius 26mm. If PA = PB and AB = 48mm, calculate the length of AP.

integral95 · Apr 7, 2015

I don't really understand the first one lol.

For the 2nd one

Let the Midpoint of AB be M

construct OM (where O is the center) and you'll see that angle OMA is a right angle as a radius bisects a chord at right angles.

From triangle OMA, use Pythagoras to find OM (i got 10 mm), you'll see that MD = OD + OM = 10+ 26 = 36

then use Pythagoras in triangle AMD to find AD (AM = 24, MD = 36) etc etc.

Feynman · Apr 7, 2015

For 1:

FX = YF (line from the centre that meets a chord at right angles bisects it)

CO = EX + FX (opposite sides in a rectangle are equal)
Then CO - FX = EX
and DO - YF = YG (similarly)
But CO = DO (radii of a circle are equal)
And FX = YF (proven above)

Therefore CO - FX = EX and CO - FX = YG
And therefore, EX = YG

InteGrand · Apr 7, 2015

Feynman said:
Now, CO - FX = EX
and DO - YF = YG

This isn't true in general. On the right track though.

InteGrand · Apr 7, 2015

abdullion said:
So I've been struggling to answer these two questions, please try help if you can!

1. CD is a diameter of the circle. CE, OF and DG are each perpendicular to the line EXFYG.(X and Y are points on the circle, and EG cuts the circle at two points.) Prove that EX = YG.

$\underline{Case 1: $X$ and $Y$ are on the same side of $CD$}$

$We know that $FX$=$FY$ (perpendicular from centre of circle to a chord bisects it). So it suffices to show that $EF$=$GF$ (note that $E$ and $G$ must lie outside the circle, so $EX=EF-FX$ and $GY=GF-FY$). Suppose that $E$ is above $G$ if we position the circle so that $CD$ lies on the $x$-axis (if $EY\parallel CD$, Feynman's proof is correct; if $G$ is above $E$, we can just flip the circle about the $y$-axis and we will effectively have $E$ above $G$).$

$From $F$, construct the line to $CE$ parallel to $CD$. Let this line hit $CE$ at the point $A$. By construction, $FA\parallel CO$, and also $FO\parallel EC\Rightarrow FO\parallel AC$. Hence $FACO$ is a parallelogram (opposite sides parallel), so $FA=CO=r$ (opposite sides of a parallelogram are equal), where $r$ is the circle's radius.$

$$Similarly, construct the point $B$ on $FO$ such that $GB\parallel DO$, and we will get that $GB=OD=r$$.$

$Note that $\angle EFA = \angle FGB=\theta$, say (this is fairly easy to prove using properties of angles in parallelograms and the facts that certain angles in the diagram are 90^\circ).$

$Then from right-angled triangle $EFA$, we get $EF=FA\cos \theta = r\cos \theta$, and similarly from triangle $FGA$, $FG=GA\cos \theta = r\cos \theta = EF$, as required. \blacksquare$

See if you can prove Case 2, where X and Y are on opposite sides of CD.

InteGrand · Apr 7, 2015

abdullion said:
So I've been struggling to answer these two questions, please try help if you can!

1. CD is a diameter of the circle. CE, OF and DG are each perpendicular to the line EXFYG.(X and Y are points on the circle, and EG cuts the circle at two points.) Prove that EX = YG.

$\underline{Case 2: $X$ and $Y$ lie on opposite sides of $CD$}$

$The proof is similar to Case 1's proof. This time, $E$, $F$, and $G$ lie inside the circle. Suppose $E$ is below $G$. Let $B$ be the point on $OF$ produced such that $GB\parallel DO$, then similarly to before, we get $GB=DO=r$. Let $A$ be the point on $FO$ produced such that $AE\parallel OC$, then again we get $AE=OC=r$. $AE\parallel OC \parallel GB\Rightarrow AE\parallel GB$, so $\angle FAE=\angle FBG$. With these equal angles, we can use right-angled trigonometry in triangles $FBG$ and $FAE$ to show that $FE=FG$, as required.$

InteGrand · Apr 7, 2015

Feynman said:
For 1:

FX = YF (line from the centre that meets a chord at right angles bisects it)

CO = EX + FX (opposite sides in a rectangle are equal)
Then CO - FX = EX
and DO - YF = YG (similarly)
But CO = DO (radii of a circle are equal)
And FX = YF (proven above)

Therefore CO - FX = EX and CO - FX = YG
And therefore, EX = YG

COFX need not be a rectangle. In the case that it is, your proof is valid.

circle geometry year 11 extension 1 questions please help! (2 Viewers)

abdullion

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integral95

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Feynman

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InteGrand

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InteGrand

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InteGrand

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