Brainliest and 20 points asap
given: circle k(o) with ot ⊥ xy, ou ⊥ wz , and ot≅ou, prove: △xoy ≅ △zow

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

a 10 percent increase