Two triangles inscribed in common circle?

Question

So, my prof hasn't really taught us much (as in, virtually nothing) about triangles inscribed in circles....

Triangles ABC and DEF are inscribed in a common circle, while angle A = angle D and angle B = angle E. Prove the triangles are congruent.

Any/some help would be greatly appreciated!

Answer 1

(A + B + C) = 180 and (D + E + F) = 180 because the sum of all the angles in a triangle total 180°.

That implies 180 - (A + B) = C and 180 - (D + E) = F. But, since A = D and B = E, 180 - (A + B) = C is equivalent to 180 - (D + E) = C. Also, since 180 - (D + E) = F, then C = F. This proves similarity of triangles by A A A (Angle, Angle, Angle).

Now, because the triangles were inscribed in the same circle, all their corresponding linear parts must be equal. This is true because if the measures of two angles inscribed in the same circle are equal, they subtend equal arcs of the circle and the chords joining the endpoints of each arc are also equal in length. That means corresponding sides opposite every pair of equal angles are equal in length. So, we have three pairs of equal angles with three pairs of equal length sides (i.e. each pair of angles are equal and each pair of sides are equal in length). Therefore, the triangles are not just similar. They are congruent.

Answer 2

I guess it depends just how much you do know.

Do you know that an angle with vertex on a circle intersects an arc of measure twice the angle (for instance, a right triangle, 90 degrees, has hypotenuse a diameter, 180 degrees of arc)

If you know that, then since angle A = angle D, then arc BC = arc EF, so chord BC = chord EF. Similarly, B = E, so AC = DF.

Now you have 2 triangles with 3 angles equal (since A + B + C = 180) and 2 sides, so ASA will show they are congruent.

Answer 3

Draw a equilateral triangle which is completelyinscribed in a circle of radius. From the center of the circle, connect the vertices using three line segments which will form three congruent isoceles triangle with congruent sides of 6cm. We know that the angle at the center is 360. Since there are three congruent isoceles triangle. The angle at vertices at each isoceles triangle is given by 360/3 = 120 degrees. Consider one of the isoceles triangle. In that the angle connecting the congruent sides is 120 degrees and the congruent sides are 6 cm. Draw perpendicular from the vertices having 120 degrees to the side opposite to that angle in that isoceles triangle. This perpendicular line acts as both angle bisector and perpendicular bisector to that Isoceles triangle. Then it will form a 30-60-90 right triangle. In which hypotenuse will be 6 cm. Let us take the side length which is opposite to 60 degrees of that right triangle as "x". The length "x" is the half the length of the equilateral triangle. Cos 30 = x/6 x = 6 cos 30 x= 6(sqrt(3)/2) x = 3sqrt(3) Length of the equilateral triangle = 2(x) = 2(3sqrt(3) ) = 6 sqrt (3) = 10.39 cm.

Answer 4

sorry, but im really having a hard time imagining the figure. Care giving more details???