1. ## Circle help needed

1. If one side of a cyclic quadrilateral is produced, prove that the exterior angle, so formed is equal to the opposite interior angle of the quadrilateral.
2. ABC is an isosceles triangle. D and E are any points of AB and AC respectively. If DE||BC (DE is parallel to BC), prove that DBCE is a cyclic quadrilateral.
3. If AC and BD are two equal chords on the opposite sides of diameter AB of a circle with centre O, prove that AC||BD.
4. Two equal circles intersect at A and B. Through A straight line MAN is drawn terminated by the circumferences show that BM=BN.

2. 1) In a cyclic quadrilateral, the sum of opposite angles is $180^{\circ}$.
Let $ABCD$ be a cyclic quadrilateral, $E\in AD$ such that $A\in (ED)$.
We have $A+C=180^{\circ}$ and $\widehat{BAD}+\widehat{EAB}=180^{\circ}$. Then $\widehat{EAB}=C$

3. 2) Let $ABC$ be an isosceles triangle with $AB=AC$. Then $\widehat{B}=\widehat{C}$.
If $DE\parallel BC$ then $\widehat{ADE}=\widehat{B}=\widehat{C}\Rightarrow \widehat{BDE}+\widehat{ECB}=180^{\circ}\Rightarrow DECB$ is cyclic.

4. 3) In triangles $ACB$ and $BDA$ we have:
$\widehat{ACB}=\widehat{BDA}=90^{\circ}$
$AB=AB$
$AC=BD$
Then $\triangle ACB\equiv\triangle BDA\Rightarrow \widehat{CAB}=\widehat{ABD}\Rightarrow AC\parallel BD$

5. 4) Let $\omega_1,\omega_2$ be the two circles, $M\in\omega_1, \ N\in\omega_2$.
$m(\widehat{AMB})=\frac{1}{2}m(arc(AB))$ and $m(\widehat{ANB})=\frac{1}{2}m(arc(AB))$.
Then $\widehat{AMB}=\widehat{ANB}\Rightarrow \triangle MBN$ isosceles $\Rightarrow MB=NB$

6. ## Too complex for me :(

I've just learnt Circle (learnt 8 theorems so far).

Please give simpler solutions. I am a Grade 10 student. Also please give reasons for what you are doing.

(I figured out the first one, please only solve the others.)

7. Actually, there're simple solutions, just take a look.