1. ## more circle

I am studying, and there are just 3 problems I don't quite understand. Plz help me.

1. Let Γ be a circle with center Q, let [AB] and [CD] be chords of Γ(so that the endpoints lie on the circle), and let G and H be the midpoints of [AB] and [CD]. Prove that d(Q, G) = d(Q, H) if and only if d(A, B) = d(C, D), and d(Q, G) < d(Q,H) if and only if d(A, B) > d(C, D).

2. Let Γ be a circle with center Q, and let L be a line containing a point X on
Γ. Prove that X is the only common point of Γand L if and only if QX is perpendicular to L. (These are the usual synthetic descriptions for the tangent line to Γat X.) [ Hint : If L also meets Γat another point Y, explain why angleQXY is acute. ]

3. Let Γ be a circle with center Q, let X be a point in the exterior of Γ, and let A and B be two points of Γwhich lie on opposite sides of QX such that XA and XB are tangent to Γin the sense of the preceding exercise. Prove that d(X, A) = d(X, B).

2. Originally Posted by jenjen
I1. Let Γ be a circle with center Q, let [AB] and [CD] be chords of Γ(so that the endpoints lie on the circle), and let G and H be the midpoints of [AB] and [CD]. Prove that d(Q, G) = d(Q, H) if and only if d(A, B) = d(C, D), and d(Q, G) < d(Q,H) if and only if d(A, B) > d(C, D).
The triangles $\displaystyle \triangle QAG$ and $\displaystyle \triangle QHC$ are right triangles and $\displaystyle QG=QH, \ QA=QC=R$.
Then $\displaystyle \triangle QAG\equiv\triangle QHC\Rightarrow AG=CH\Rightarrow AB=CD$.

We have $\displaystyle R^2=QA^2=QG^2+AG^2$
$\displaystyle R^2=QC^2=QH^2+CH^2$.
Then:
$\displaystyle QG<QH\Rightarrow QG^2<QH^2\Rightarrow QG^2-QH^2<0$
$\displaystyle QG^2-QH^2=(R^2-AG^2)-(R^2-CH^2)=CH^2-AG^2$
Then $\displaystyle CH^2<AG^2\Rightarrow CH<AG\Rightarrow 2CH<2AG\Rightarrow CD<AB$

3. Originally Posted by jenjen
I am studying, and there are just 3 problems I don't quite understand. Plz help me.

2. Let Γ be a circle with center Q, and let L be a line containing a point X on
Γ. Prove that X is the only common point of Γand L if and only if QX is perpendicular to L. (These are the usual synthetic descriptions for the tangent line to Γat X.) [ Hint : If L also meets Γat another point Y, explain why angleQXY is acute. ]
.
X is the only common point of $\displaystyle \Gamma$ and L

$\displaystyle \Longleftrightarrow$ L is tangent to $\displaystyle \Gamma$

$\displaystyle \Longleftrightarrow$ the line connecting X and the center of $\displaystyle \Gamma$ is perpendicular to L..

if we use the hint..
suppose L meets $\displaystyle \Gamma$ at points X and Y.. then $\displaystyle \Delta QXY$ is an isosceles triangle with $\displaystyle \overline {QX}=\overline {QY}$. hence, $\displaystyle m\angle QXY = m\angle QYX$
now, if $\displaystyle m\angle QXY \geq 90^o$, then the sum of the internal angles of $\displaystyle \Delta QXY$ would be > $\displaystyle 180^o$

therefore, $\displaystyle m\angle QXY < 90^o$.. QED

4. Originally Posted by jenjen
3. Let Γ be a circle with center Q, let X be a point in the exterior of Γ, and let A and B be two points of Γwhich lie on opposite sides of QX such that XA and XB are tangent to Γin the sense of the preceding exercise. Prove that d(X, A) = d(X, B).
consider $\displaystyle \triangle QAX$ and $\displaystyle \triangle QBX$

XA and XB are tangent lines, $\displaystyle \implies m\angle QAX = 90^o = m\angle QBX$.. also, $\displaystyle \overline {QA} = \overline {QB}$ and $\displaystyle \overline {QX} = \overline {QX}$.. hence, $\displaystyle \triangle QAX \sim \triangle QBX$. therefore, $\displaystyle \overline {XA} = \overline {XB}$. QED

5. Hello, jenjen!

Here's #3 . . .

3. Let C be a circle with center Q.
Let X be a point in the exterior of C,
and let A and B be two points of C which lie on opposite sides of QX
such that XA and XB are tangent to C.
Prove that: .$\displaystyle XA = XB$
Code:
              * * *
*           *  A
*               *
*              /  * *
/         *
*           /       *     *
*        Q* - - - - * - - - *X
*           \       *     *
\         *
*              \  * *
*               *
*           *  B
* * *

Tangent XA is perpendicular to radius QA.
Tangent XB is perpendicular to radius QB.
. . Hence, triangles QAX and QBX are right triangles.

Since QA and QB are radii of circle Q, $\displaystyle QA = QB.$
. . And, of course: .$\displaystyle QX = QX.$

Hence: .$\displaystyle \Delta QAX \cong \Delta QBX$

Therefore: .$\displaystyle XA = XB$

6. thank you sooo much!