:
$\text{In}\mathrm{Δ}AOC\text{and}\mathrm{Δ}AOB,OC=OB\text{(Given)}\mathrm{âˆ}OBC=\mathrm{âˆ}OCA\text{(Both angles are right angles)}\text{And}AO\text{is common side}.\text{Hence,}\mathrm{Δ}AOCâ‰\dots \mathrm{Δ}AOB\text{(RHS congruence)}∴\mathrm{âˆ}CAO=\mathrm{âˆ}BAO\text{(CPCT)}⇒2x+24=30⇒x=3⇒x−3=0$

Answer: Option C. -> ∠NVY+  2∠VYN = 180∘
:
C
Given, in $\mathrm{â–³}NVY$, NV = VY.

Then, $\mathrm{âˆ}VYN=\mathrm{âˆ}YNV$.
[angles opposite to equal sides of a triangle are equal]
Using angle sum property of a triangle, we get:
$\mathrm{âˆ}NVY+\mathrm{âˆ}VYN+\mathrm{âˆ}YNV={180}^{∘}$
So, $\mathrm{âˆ}NVY+2\mathrm{âˆ}VYN={180}^{∘}$.
It is also given that $VYâ‰YN$. Hence the options $\mathrm{âˆ}NVY=\mathrm{âˆ}VYN$ and$\mathrm{âˆ}NVY=\mathrm{âˆ}YNV$ are not correct.

Answer: Option C. -> Statement 1 is correct but 2 is false
:
C
$\text{In}\mathrm{Δ}APB\text{and}\mathrm{Δ}CQD,AB=CD\text{(opposite sides of a parallelogram)}.\mathrm{âˆ}ABP=\mathrm{âˆ}CDQ\text{(alternate angles)}.\mathrm{âˆ}APB=\mathrm{âˆ}CQD\text{(both are right angles)}.∴\mathrm{Δ}APBâ‰\dots \mathrm{Δ}CQD$
(using AAS criterion)
Hence, statement 1 is correct but statement 2 is false

Answer: Option B. -> False
:
B
The given statement is false. Even when two triangles have all angles same, they can still have sides of different lengths. However, the ratio of lengths of corresponding sides will be same. But in this case, they will be called 'similar' triangles, not congruent.
Congruent triangles are a special case of similar triangles.
For example, in the image below, both triangles are equilateral and have all angles equal but they are not congruent.

Answer: Option C. -> BC
:
C

$\mathrm{âˆ}A+\mathrm{âˆ}B+\mathrm{âˆ}C={180}^{∘}$ (angle sum property of triangle)
$⇒\mathrm{âˆ}A+{45}^{∘}+{45}^{∘}={180}^{∘}⇒\mathrm{âˆ}A={90}^{∘}$
The side opposite to largest angle will be the longest side. Hence BC is the largest side.

Answer: Option A. -> True
:
A
Since $\mathrm{Δ}ABCâ‰\dots \mathrm{Δ}BED,\mathrm{âˆ}DBE=\mathrm{âˆ}CAB={25}^{∘}\text{(CPCT)}\text{And,}\mathrm{âˆ}DEB=\mathrm{âˆ}CBA\text{(CPCT)}\text{Now,}\mathrm{âˆ}BDE+\mathrm{âˆ}DBE+\mathrm{âˆ}BED={180}^{∘}⇒\mathrm{âˆ}BDE+{25}^{∘}+{115}^{∘}={180}^{∘}⇒\mathrm{âˆ}BDE={40}^{∘}\text{Also,}\mathrm{âˆ}CBA+\mathrm{âˆ}CBD+\mathrm{âˆ}DBE={180}^{∘}\text{(Linear set of angles)}⇒\mathrm{âˆ}CBA+{25}^{∘}+{115}^{∘}={180}^{∘}⇒\mathrm{âˆ}CBA={40}^{∘}⇒\mathrm{âˆ}BDE=\mathrm{âˆ}CBD\text{and they are alternate angles, so}BC||ED.$

:

In $\mathrm{â–³}$AOB and$\mathrm{â–³}$DOC,
OB= OC (Radii of the same circle)
OA = OD (Radii of the same circle)
AB = CD (Given)
$\mathrm{â–³}$AOB $â‰\dots$$\mathrm{â–³}$DOC (By SSS congruence condition)
So, $\mathrm{âˆ}$ABO = $\mathrm{âˆ}$DCO = ${35}^{∘}$(Corresponding parts of congruent triangles)

Answer: Option C. -> 16∘
:
C
In given figure,
$\mathrm{âˆ}BAC+\mathrm{âˆ}ABC+\mathrm{âˆ}ACB={180}^{∘}\text{(Angles of same triangle)}⇒\mathrm{âˆ}BAC+{65}^{∘}+{33}^{∘}={180}^{∘}⇒\mathrm{âˆ}BAC={82}^{∘}$
$\mathrm{âˆ}CAN=\frac{\mathrm{âˆ}BAC}{2}={41}^{∘}\text{(Given that AN is an angle bisector)}.\text{Now,}\mathrm{âˆ}CAN+\mathrm{âˆ}ANC+\mathrm{âˆ}ACN={180}^{∘}\text{(Angles of same triangle)}⇒{41}^{∘}+\mathrm{âˆ}ANC+{33}^{∘}={180}^{∘}⇒\mathrm{âˆ}ANC={106}^{∘}$
$\mathrm{âˆ}ANC+\mathrm{âˆ}ANM={180}^{∘}\text{(Linear pair)}⇒{106}^{∘}+\mathrm{âˆ}ANM={180}^{∘}⇒\mathrm{âˆ}ANM={74}^{∘}$
$\mathrm{âˆ}ANM+\mathrm{âˆ}AMN+\mathrm{âˆ}MAN={180}^{∘}⇒{74}^{∘}+{90}^{∘}+\mathrm{âˆ}MAN={180}^{∘}⇒\mathrm{âˆ}MAN={16}^{∘}.$

Answer: Option A. -> True
:
A

In $\mathrm{â–³}$ABC and $\mathrm{â–³}$CDA, we have:
AB = CD (Opposite sides of a parallelogram)
BC = AD (Opposite sides of a parallelogram)
AC = AC (Common)
By SSS congruence condition, $\mathrm{â–³}$ABC$â‰\dots$$\mathrm{â–³}$CDA. So, the diagonal of the parallelogram divides it into two congruent triangles.

Answer: Option B. -> Isosceles but not right angled
:
B
$\mathrm{âˆ}$ POY = $\mathrm{âˆ}$ OYZ(alternate angles)
$\mathrm{âˆ}$Y is bisected, so$\mathrm{âˆ}$ POY =$\mathrm{âˆ}$ PYO
Hence, PY = PO
So,$\mathrm{â–³}$YPO is isosceles
Also, it is given that$\mathrm{âˆ}$XYZ is acute, so any angle which is half of it (bisected by OY) is less than ${45}^{∘}$.
or$\mathrm{âˆ}$ PYO +$\mathrm{âˆ}$OYZ < ${90}^{∘}$
Hence, the third angle of the$\mathrm{â–³}$YPO i.e.$\mathrm{âˆ}$ YPO will be obtuse to satisfy angle-sum property of a triangle.
Hence,$\mathrm{â–³}$YPO is isosceles but not rightangled.