Intersecting Circles

We have circles $\Gamma_1$ and $\Gamma_2$ intersect at points B and C
$\Rightarrow$ B and C are symmetrical through the line $O_1O_2$ .
Therefore, $O_1O_2$ is the bisector of $\angle BO_1C$ ,
$\Rightarrow \angle BO_1A = 26 ^ \circ$ .

We have: 4 points $B, E, A, O_1$ on the circumference of $\Gamma_2$ .
$\Rightarrow BEAO_1$ is a cyclic quadrilateral of circle $\Gamma_2$ .
$\Rightarrow$ the acute angle between lines $DE$ and $EA$ is equal to $\angle BO_1A$ ( because the exterior angle is equal to the interior opposite angle)
$\Rightarrow$ the acute angle between lines DE and EA is equal to $26^ \circ$ .

This question is looking for the acute angle between $DE$ and $EA$ , so after drawing a diagram it is clear that that refers to $\angle AEB$

Firstly since $CO'$ and $O'B$ are radii of $\Gamma_1$ , $\bigtriangleup CO'B$ is isoceles. $O'A$ is the angle bisector of angle $CO'A$ , so $\angle AO'B=26^\circ$

Join $AB$ . We see that $\angle AEB$ and $\angle AO'B$ share a common base $AB$ . Because angles on the same segment are equal, that means that $\angle AEB=26^\circ$ . Thus, the answer is $26$

Draw the picture, assume that the radius of $\Gamma_1$ is $R$ , and the radius of $\Gamma_2$ is $r$

and see that the $\triangle BO_2O_1$ and $\triangle CO_2O_1$ , we obtain that : $r = O_2C = O_2B$ , $O_2O_1 = O_2O_1$ , and $R = O_1C = O_1B$

Then, we can conclude that $\triangle BO_2O_1 \equiv \triangle CO_2O_1$ . So, we obtain :

$\angle BO_1A = \angle AO_1C = \frac{1}{2} (\angle BO_1C = \frac{52^{\circ}}{2} = 26^{\circ}$

As we know that, $\angle BEA = \angle BO_1A = \boxed{26^{\circ}}$ .

Then, the measure (in degrees) of the acute angle between lines $DE$ and $EA$ is $\boxed{26^{\circ}}$

Since $B$ and $C$ are symmetric about line $O_1 O_2 A$ , it follows that $\angle BO_1 A = \angle CO_1 A = 26 ^\circ$ . By circle properties (angle in the same segment), $\angle BEA = \angle BO_1 A$

Thus, $\angle DEA = 180^\circ - \angle BEA =154 ^\circ$ . Hence, the acute angle between these lines is $26 ^\circ$ .

Note: There are several ways for the diagram to be drawn, hence the question just asks for the acute angle. The result is independent of $R_2> R_1$ , where the point $D$ is, and whether line segment $DB$ intersects $E$ .

By drawing the picture, it's easy to see that $\angle{AEB} = \angle{AO_1B}$ and $\angle{AO_1B} = \angle{AO_1C} = 26^{\circ}$ . Hence, $\angle{AEB} = 26^{\circ}$

If one sets E on the center of circle one, then one must draw out the line CE. Let the point Q be where CE intersects circle one and let P be where the continuation of AE intersects circle one. The angle DEQ is equal to 52 degrees because it is the opposite angle of CEB. To find the acute angle between the Lines DE and EA, we must find the angle between EP and DE. Since EP bisects DEQ, then the angle is 52/2. Thus the answer is 26 degrees.

From the question, we want to find the acute angle of intersection between lines $DE$ and $EA$ . Since $DB$ is simply an extension of the line segment $DE$ , we can rephrase the purpose of the question to be to find the acute angle of intersection between lines $DB$ and $EA$ , which is $\angle BEA$ .

Following on, since $\angle BEA$ comes from the same sector as $\angle ACB$ , we can conclude that $\angle BEA = \angle ACB$ .

From the start of the question, one will notice that triangle $O_1BC$ is an isosceles triangle, with $O_1B=O_1C=R_1$ . This means that $\angle O_1CB = \angle O_1BC = \frac {180^\circ - 52^\circ}{2} = 64^\circ$ , being the base angles of an isosceles triangle.

Also, as $O_1$ lies on $\Gamma_2$ , $O_1O_2 = R_2$ . Since, $O_1O_2$ intersect again at $A$ on $\Gamma_2$ , this implies that $O_1A$ is the diameter of $\Gamma_2$ .

Thus, $\angle O_1CA = 90^\circ$ as it is the angle formed in a semicircle.

From this, we find that $\angle ACB=\angle O_1CA - \angle O_1CB = 90^\circ - 64^\circ = 26^\circ$ .

Hence, $\angle ACB = \angle AEB = 26^\circ$ .

Thus, the measure of the acute angle between $DE$ and $EA$ is $26^\circ$ .