ΔXOY ≅ ΔZOW ⇒ proved down Step-by-step explanation: * Lets study some facts on the circle - If two chords equidistant from the center of the circles, then they are equal in length * the meaning of equidistant is the perpendicular distances from the center of the circle to the chords are equal in length * Lets check this fact in our problem ∵ XY and WZ are two chords in circle O ∵ OT ⊥ XY - OT is the perpendicular distance from the center to the chord XY ∵ OU ⊥ WZ - OU is the perpendicular distance from the center to the chord WZ ∵ OT ≅ OU - The two chords equidistant from the center of the circle ∴ The two chords are equal in length ∴ XY ≅ WZ * Now in the two triangles XOY and ZOW , to prove that they are congruent we must find one of these cases: 1- SSS ⇒ the 3 sides of the 1st triangle equal the corresponding sides in the 2nd triangle 2- SAS ⇒ the two sides and the including angle between them in the 1st triangle equal to the corresponding sides and including angle in the 2nd triangle 3- AAS ⇒ the two angles and one side in the 1st triangle equal the corresponding angles and side in the 2nd triangle * Lets check we will use which case - In the two triangles XOY and ZOW ∵ XY = ZW ⇒ proved ∵ OX = OZ ⇒ radii ∵ OY = OW ⇒ radii * This is the first case SSS ∴ ΔXOY ≅ ΔZOW