Question

Question in attachment.

Answer 1

Given -
AB and CD are chords of a circle which intersect at right angles.

To Prove - arc CXA + arc DZB = arc AYD + arc BWC

Construction -

Join OA , OC , OD, OB , AD and CB

Let angle BCD be x.

Therefore angle DOB = 2x ( angle subtended by a segment at center is double the angle subtended by it at circumference; In this case DB subtends x on circumference so it'll subtend 2x at center)

Since angle BCD = x
angle ABC = 180-(90+x) = 90-x ( angle sum property of triangle)

angle AOC = 2(angle ABC) [ angle subtended at the center is double the angle subtended at arc)
→ angle AOC = 2(90-x) = 180-2x

Therefore, angle AOC + angle DOB = 180-2x + 2x = 180°

→ angle AOC + angle DOB = 180°

Hence , angle AOD + angle COB = 360 - (angle AOC + angle DOB)
→ angle AOD + angle COB = 360 - 180° = 180°

Now since :

angle AOC + angle DOB = angle AOD + angle COB
→ arc CXA + arc DZB = arc AYZ + arc BWC
[ since sum of the angles subtended by the arcs is equal , length of arcs will also be equal]

Answer 2

Given that 2 chords AB and CD of a circle intersect at right angles.


In triangle AOD:

We know that Sum of all angles of triangle = 180.

DOA + ODA + AOD = 180

DOA + ODA + 90 = 180

DOA + ODA = 180 - 90

DOA + ODA  = 90  ---- (1)

Now, angle DOA & ODA are subtended by arc DZB & arc CXA respectively.

arc DZB + arc CXA = 90  ------ (2).


In triangle DOB:

BOD + OBD + DOB = 180

BDO + OBD + 90 = 180

BDO + OBD = 90  -   ------ (3)

Now, angle BDO and OBD are subtended by arc BWC & arc AYD respectively.

arc BWC + arc AYD = 90   ------- (4)


From (3) & (4), 

arc DZB + arc CXA = arc BWC + arc AYD = 90.

arc DZB + arc CXA = arc BWC + arc AYD = semicircle.


Note: Sorry, I didn't solve it on my own. 


Hope this helps!