Ans:- Construction:-
1) join MX
2) join YM
3) join MP
4) join XP
5) join YP
6) join MP
7)Let the mid point of side MX be L.
8) Let the mid point of side YN be K.
9) Let the intersection point of chords MN and XY be S.
10) Draw a perpendicular from point P to the mid point of side MX passing through the mid point of YN and the centre of the circle.


Solution:- In ∆ MPL and ∆XPL,

ML = XL...
(perpendicular drawn from the centre of the circle to a chord bisects the chord)

∠MLP = ∠XLP ...
(Construction => each 90)

LP = LP...
(Common)

Therefore,
(∆MPL congruent ∆XPL .By SAS congruence rule).

=> 1) MP = XP

Now,In ∆YPK and ∆NPK,

YK = NK..
(Perpendicular drawn from the centre of the circle to the chord bisects the chord)

∠YKP = ∠NKP ..
(Construction => each 90)

KP = KP...
(Common)

Therefore,

(∆YPK is congruent to ∆MPK.By SAS congruence rule)

=> 2) YP = NP .

Hence proved.





NOTE from Me:-

1)Some might think why is YN perpendicularly bisected it is because PL is perpendicular to MX while passing through the centre of the circle and also the mid point of YN therefore it will cause the SN to be perpendicularly bisect YN you will understand this clearly if you see the figure)
2) My question :- Can we do a construction like this in exams ? it is quite long.Is a long construction acceptable?
3) If there is some mistake please tell me.

Proof:

Given:
- MN and XY are two equal chords of a circle with center ‘O’.
- MN and XY, when produced, meet at external point ‘P’ outside the circle.

Proof:

Step 1:
- Draw a diagram of the circle with center O, and the chords MN and XY intersecting at point P outside the circle.

Step 2:
- Since MN and XY are equal chords, they are equidistant from the center O of the circle.
- This implies that OP is the perpendicular bisector of MN and XY.

Step 3:
- In triangle NPO and triangle YPO,
- OP is common,
- NP = YP (as they are equal chords),
- ∠NPO = ∠YPO = 90° (as OP is perpendicular to both MN and XY).

Step 4:
- By SAS (side-angle-side) congruence criterion,
- triangle NPO ≅ triangle YPO.

Step 5:
- Corresponding parts of congruent triangles are equal,
- NP = YP.

Step 6:
- Similarly, by considering triangles MPO and XPO,
- MP = XP.

Step 7:
- Therefore, we have proved that NP = YP and MP = XP when MN and XY are two equal chords of a circle with center ‘O’ meeting at external point ‘P’ outside the circle.