Given: The side BC of a ∆ABC is produced to D. The bisector of ∠BAC intersects the side BC at E.
To Prove: ∠ABC + ∠ACD = 2 ∠AEC.
Proof: In ∆ABE,
∠AEC = ∠ABC + ∠BAE
| Exterior angle theorem
= ∠ABC + ∠CAE ...(1)
| ∠BAE = ∠CAE (∵ AE bisects ∠BAC)
In ∆AEC,
∠ACD = ∠AEC + ∠CAE
| Exterior angle theorem
⇒ ∠CAE = ∠ACD - ∠AEC    ...(2)
From (1) and (2),
∠AEC = ∠ABC + (∠ACD - ∠AEC)
⇒ 2 ∠AEC = ∠ABC + ∠ACD

Given: ABCD is a quadrilateral
To Prove: ∠A + ∠B + ∠C + ∠D = 360°
Construction: Join AC.
Proof: In ∆ABC,


∠1 + ∠B + ∠3 = 180°    ...(1)
| Angle sum property of a triangle
In ∆ADC,
∠2 + ∠D + ∠4 = 180°    ...(2)
| Angle sum property of a triangle
Adding (1) and (2), we get
(∠1 + ∠2) + ∠B + (∠3 + ∠4) + ∠D = 360°
∠    ∠A + ∠B + ∠C + ∠D = 360°

Given: PS is the bisector of ∠PQR and PT⊥QR.
To Prove:   

Proof: ∵ PS is the bisector of ∠QPR

∴ ∠QPS = ∠RPS
⇒ ∠1 + ∠TPS = ∠2    ...(1)
In ∆PQT,
∠PTQ = 90°    | Given
∴ ∠1 + ∠Q = 90°
| Angle sum property of a triangle
⇒    ∠Q = 90° - ∠1    ...(2)
In ∆PRT,
∠PTR = 90°    | Given
∴ ∠R + ∠TPR = 90°
| Angle sum property of a triangle
⇒ ∠R + (∠TPS + ∠2) = 90°    ...(3)
From (2) and (3),
∠Q = ∠R + (∠TPS + ∠2) - ∠1
⇒ ∠Q - ∠R = ∠TPS + (∠2 - ∠1)
⇒ ∠Q - ∠R = ∠TPS + ∠TPS | From (1) ⇒ ∠Q - ∠R = 2 ∠TPS

Given: The side BC of ∆ABC is produced to D. The bisector of ∠A meets BC in L.
To Prove: ∠ABC + ∠ACD = 2 ∠ALC.
Proof: ∠ABC + ∠ACD
= ∠ABC + (∠ABC + ∠BAC)
| Exterior angle theorem
= 2 ∠ABC + ∠BAC
= 2 ∠ABC + 2 ∠BAL
| ∵ AL is the bisector of ∠A
= 2 (∠ABC + ∠BAL)
= 2 ∠ALC | Exterior angle theorem

Given: ABC is a triangle right angled at A. AL is drawn perpendicular to BC.
To Prove: ∠BAL = ∠ACB

Proof: In triangle ALB,
∠ALB + ∠BAL + ∠ABL = 180°
| Angle sum property of a triangle
⇒ 90° + ∠BAL + ∠ABC = 180°
⇒     ∠BAL + ∠ABC = 90° ...(1)
In triangle ABC,
∠BAC + ∠ACB + ∠ABC = 180°
| Angle sum property of a triangle
⇒ 90° + ∠ACB + ∠ABC = 180°
⇒ ∠ACB + ∠ABC = 90° ...(2)
From (1) and (2),
∠BAL + ∠ABC = ∠ACB + ∠ABC
⇒ ∠BAL = ∠ACB