From above given figure , we can see that
CE || BA and AC is the transversal.
∴ ∠4 = ∠1 (alternate interior ∠s) .......... ( 1 )
Again, CE || BA and BD is the transversal.
∴ ∠5 = ∠2 (corresponding ∠s) ................( 2 )

Correct Option: C

From above given figure , we can see that
CE || BA and AC is the transversal.
∴ ∠4 = ∠1 (alternate interior ∠s) .......... ( 1 )
Again, CE || BA and BD is the transversal.
∴ ∠5 = ∠2 (corresponding ∠s) ................( 2 )
Adding equations ( 1 ) and ( 2 )
∴ ∠4 + ∠5 = ∠1 + ∠2
∴∠ACD = ∠A + ∠B.

As per given figure , we can see that
In ΔBCE, BC = EC, ∴ ∠B = ∠BEC
In ΔCDE, ED = EC, ∴ ∠ECD = ∠EDC
and In ΔADE, AD = ED, ∴ ∠AED = ∠A
Now ∠B = ∠BEC = ∠A + ∠ECD = ∠A + ∠EDC = ∠A + ∠EAD + ∠AED
{ ∴ ∠EDC = ∠EAD + ∠AED }
=∠A + ∠A + ∠A = 3∠A { By ASA property }

Correct Option: D

As per given figure , we can see that
In ΔBCE, BC = EC, ∴ ∠B = ∠BEC
In ΔCDE, ED = EC, ∴ ∠ECD = ∠EDC
and In ΔADE, AD = ED, ∴ ∠AED = ∠A
Now ∠B = ∠BEC = ∠A + ∠ECD = ∠A + ∠EDC = ∠A + ∠EAD + ∠AED
{ ∴ ∠EDC = ∠EAD + ∠AED }
=∠A + ∠A + ∠A = 3∠A { By ASA property }

PQRS is a square .
We know that Every angle of square is 90° and the diagonals bisect each other at right angles .
SP = SR and ∠S = 90°

Correct Option: A

PQRS is a square .
We know that Every angle of square is 90° and the diagonals bisect each other at right angles .
SP = SR and ∠S = 90°

If the in circle of a triangle ABC touches BC at D, then |BC - CD| = |AB - AC|
In our case, AC = 5, AB = 3
⇒ AC – AB = 2
∴ CD – BD = 2

Correct Option: C

First method :-
If the in circle of a triangle ABC touches BC at D, then |BC - CD| = |AB - AC|
In our case, AC = 5, AB = 3
⇒ AC – AB = 2
∴ CD – BD = 2
In our case, BC = 4
⇒ BD + BD = 4 and – BD + DC = 22
⇒ CD = 3
⇒ BD = 1 = OE = Radius of the in circle.
second method :-
We know that , In right angle triangle
Here , a = 3 cm , b = 4 cm and c = 5 cm
We know that ,

Let ABCD is a rhombus in which AB = BC = CD = DA = diagonal CA.
In triangle ABC ,
AB = BC = CA , therefore ∠B = 60°
∠D = ∠B = 60°
∠A=180 - ∠B = 180° - 60° =120°
∠C = ∠A=120°
Hence ,The ratio of diagonals = √2 : 1

Correct Option: B

Let ABCD is a rhombus in which AB = BC = CD = DA = diagonal CA.
In triangle ABC ,
AB = BC = CA , therefore ∠B = 60°
∠D = ∠B = 60°
∠A=180 - ∠B = 180° - 60° =120°
∠C = ∠A=120°
Hence ,The ratio of diagonals = √2 : 1