Answer:
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
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