1. ## Circle Geometry Questions

Most circle geometry questions I understand, but these are from my homework that I don't understand.

2. Hello, skitty!

Here's the first one . . .

[quote]1. Prove that the triangles in the diagram are both right triangles.
Code:
              Q
o * *
*     *     *
*   *     *     *
*            *    *
*            *
*                 * *
P o * * * * * * * * * o R
*                 * *
*            *
*            *    *
*   *     *     *
*     *     *
o * *
S

We are given: .$\displaystyle PQ = PS,\;\;RQ = RS$

Since $\displaystyle PR = PR,\;\Delta PQR \cong \Delta PSR\;\;\text{ (s.s.s.)}$

Then: $\displaystyle \angle QPR = \angle SPR\quad\Rightarrow\quad \text{arc(QR)} = \text{arc(SR)}$ .[1]

And: $\displaystyle \angle QRP = \angle SRP\quad\Rightarrow\quad \text{arc(PQ)} = \text{arc(PS)}$ .[2]

Add [1] and [2]: .$\displaystyle \text{arc(PQ) + arc(QR)} \:=\:\text{arc(PS) + arc(SR)}$

. . That is: .$\displaystyle \text{arc(PQR)} \:=\:\text{arc(PSR)}$

Since $\displaystyle \text{arc(PQR)} + \text{arc(PSR)} \:=\:360^o$, then: .$\displaystyle \text{arc(PQR)} \:=\:\text{arc(PSR)} \:=\:180^o$

Since $\displaystyle \angle Q\text{ and }\angle S$ are inscribed in semicircles, $\displaystyle \angle Q \:=\:\angle S \:=\:90^o$

3. ## dear skitty

HELLO SKITTY

TO PROVE-
PERIMETER (PQR)=2PA
PROOF-
PA=PB (Tangents from external point)
Also;
QX=BQ and AR=RX (Tangents from external point)
now;
PQ+QX=PR+XR
PQ+QX+PR+XR=PERIMETER(PQR)
2(PR+RX)=Perimeter(PQR)
2PA=Perimeter (PQR)
hence proved.

4. ## dear skitty

hello skitty;

TO PROVE-
angle QOS=90

PROOF-
anglePQS+angleRSQ=180 (CO-INTERIOR ANGLES)

anglePQO=angleSQO and angleRSO=angleQSO

NOW;
1/2 (anglePQS+angleRSQ)=90
angleSQO+angleQSO=90

BY ANGLE SUM PROPERTY OF TRIANGLE-
angleSQO+angleQSO+ angleQOS=180
angleQOS=180-90=90
So angleQOS=90

HENCE PROVED