Solution:

Point A lies on the bisector of \(\angle\)XYZ. ... (Given)
Every point on the bisector of an angle is equidistant from the sides of the angle. ... (Theorem)
∴ AZ = AX
But, AX = 2 cm ... (Given)
∴ AZ = 2 cm

Solution:

seg PT \(\perp\) ray ST ... (Given)
seg PR \(\perp\) ray SR ... (Given) and
seg PR \(\cong\) seg PT ... (Given)
∴ Point P is equidistant from the sides of \(\angle\)RST.

Any point equidistant from the sides of an angle, lies on the bisector of the angle. ... (Theorem of angle bisector)
∴ Ray SP is the bisector of \(\angle\)RST.

∴ \(\angle\)RSP = \(\displaystyle \frac{1}{2}\) \(\angle\)RST

∴ \(\angle\)RSP = \(\displaystyle \frac{1}{2}\) × 56° ... (Given)

∴ \(\angle\)RSP = 28°

Solution:

If two sides of a triangle are unequal, then the angle opposite to the greater side is greater than angle opposite to the smaller side. ... (Theorem)

Now, 12 > 10 > 8

∴ QR > PQ > PR

∴ \(\angle\)P > \(\angle\)R > \(\angle\)Q

∴ \(\angle\)P is the greatest angle and \(\angle\)Q is the smallest angle of \(\triangle\)PQR.

Solution:

In \(\triangle\)FAN,

 \(\angle\)F + \(\angle\)A + \(\angle\)N = 180° ... (Theorem)

∴ 80° + 40° + \(\angle\)N = 180°

∴ 120° + \(\angle\)N = 180°

∴ \(\angle\)N = 180° - 120°

∴ \(\angle\)N = 60°

Now, 80° > 60° > 40°

∴ \(\angle\)F > \(\angle\)N > \(\angle\)A

∴ Side AN > Side FA > Side FN

∴ Side AN is the greatest side and side FN is the smallest side of \(\triangle\)FAN.

Given:

\(\triangle\)ABC is an equilateral triangle.
i.e. AB = BC = CA

To Prove:

\(\angle\)A \(\cong\) \(\angle\)B \(\cong\) \(\angle\)C
i.e. \(\triangle\)ABC is equiangular triangle.

Proof:

In \(\triangle\)ABC,
 BC = CA ... (Given)
∴ \(\angle\)A \(\cong\) \(\angle\)B ... (Theorem of Isosceles Triangle) ... (i)

Also, CA = AB ... (Given)
∴ \(\angle\)B \(\cong\) \(\angle\)C ... (Theorem of Isosceles Triangle) ... (ii)

From (i) and (ii), we have:
 \(\angle\)A \(\cong\) \(\angle\)B \(\cong\) \(\angle\)C ... (Transitivity)

∴ \(\triangle\)ABC is equiangular triangle.

∴ An equilateral triangle is equiangular.

Given:

In \(\triangle\)ABC, seg AD is the bisector of \(\angle\)BAC and seg AD \(\perp\) side BC.

To Prove:

seg AB \(\cong\) seg AC
i.e. \(\triangle\)ABC is an isosceles triangle.

Proof:

In \(\triangle\)ADB and \(\triangle\)ADC,
 \(\angle\)ADB \(\cong\) \(\angle\)ADC ... (90° each) ... (Given)
 seg AD \(\cong\) seg AD ... (Reflexivity)
 \(\angle\)BAD \(\cong\) \(\angle\)CAD ... (Given)
∴ \(\triangle\)ADB \(\cong\) \(\triangle\)ADC ... (A S A test)
∴ seg AB \(\cong\) seg AC ... (c s c t)
i.e. \(\triangle\)ABC is an isosceles triangle.

Proof:

In \(\triangle\)PQR,
 side PQ \(\cong\) side PR ... (Given)
∴ \(\angle\)Q \(\cong\) \(\angle\)PRQ ... (Theorem of Isosceles Triangle) ... (i)

Now, \(\angle\)PRQ is an exterior angle of \(\triangle\)PRS.
∴ \(\angle\)PRQ > \(\angle\)S ... (Theorem of exterior angle of a triangle) ... (ii)

From (i) and (ii), we have:
 \(\angle\)Q > \(\angle\)S ... (iii)

∴ In \(\triangle\)PSQ,
∴ seg PS > seg PQ ... (Theorem)

Proof:

In \(\triangle\)ADB and \(\triangle\)BEA,
 \(\angle\)ADB \(\cong\) \(\angle\)BEA ... (90° each) ... (Given)
 seg AB \(\cong\) seg BA ... (Reflexivity)
 seg BD \(\cong\) seg AE ... (Given)
∴ \(\triangle\)ADB \(\cong\) \(\triangle\)BEA ... (Hypotenuse Side Theorem)
∴ seg AD \(\cong\) seg BE ... (c s c t)