Given: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB.
To Prove: (i) ∆DAP ≅ ∆EBP
(ii) AD = BE.
Proof: (i) In ∆DAP and ∆EBP,
AP = BP
| ∵ P is the mid-point of the line segment AB
∠DAP = ∠EBP    | Given
∠EPA = ∠DPB    | Given
⇒ ∠EPA + ∠EPD = ∠EPD + ∠DPB
| Adding ∠EPD to both sides
⇒    ∠APD = ∠BPE
∴ ∠DAP ≅ ∠EBP    | ASA Rule
(ii) ∵ ∆DAP ≅ AEBP    | From (i) above
∴ AD = BE.    | C.P.C.T.

Given: In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O.
To Prove: (i) OB = OC
(ii) AO bisects ∠A.
Proof: (i) AB = AC    | Given
∴ ∠B = ∠C
| Angles opposite to equal sides of a triangle are equal





∴ ∠OBC = ∠OCB
| ∵ BO and CO are the bisectors of ∠B and ∠C respectively
∴ OB = OC
| Sides opposite to equal angles of a triangle are equal
(ii) In ∆OAB and ∆OAC,
AB = AC    | Given
OB = OC | Proved in (i) above
OA = OA    | Common
∴ ∠B = ∠C
| Angles opposite to equal sides of a triangle are equal

∴ ∠ABO = ∠ACO
| ∵ BO and CO are the bisectors of ∠B and ∠C respectively
∴ ∆OAB ≅ ∆OAC | By SAS Rule
∴ ∠OAB = ∠OAC    | C.P.C.T.
∴ AO bisects ∠A.

Given: In ∆ ABC, AD is the perpendicular bisector of BC.
To Prove: A ABC is an isosceles triangle in which AB = AC.
Proof: In ∆ ADB and ∆ADC,
∠ADB = ∠ADC    | Each = 90° DB = DC
| ∵ AD is the perpendicular bisector of BC
AD = AD    | Common
∴ ∆DB ≅ ∆ADC    | By SAS Rule
∴ AB = AC    | C.P.C.T.
∴ ∆ABC is an isosceles triangle in which AB = AC.

Given: In figure, AC = AE, AB = AD and ∠BAD = ∠EAC.
To Prove: BC = DE.
Proof: In ∆ABC and ∆ADE,
AB = AD    | Given
AC = AE    | Given
∠BAD = ∠EAC    | Given
⇒ ∠BAD + ∠DAC = ∠DAC + ∠EAC
| Adding ∠DAC to both sides
⇒    ∠BAC = ∠DAE
∴ ∆ABC ≅ ∆ADE | SAS Rule
∴ BC = DE.    | C.P.C.T.

Given: In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B.

To Prove: (i) ∆AMC ≅ ∆BMD
(ii)    ∠DBC is a right angle
(iii)    ∠DBC ≅ ∆ACB

(i) In ∆AMC and ∆BMD,
AM = BM
| ∵ M is the mid-point of the hypotenuse AB
CM = DM    | Given
∠AMC = ∠BMD
| Vertically Opposite Angles
∴ ∆AMC ≅ ∆BMD. | SAS Rule
(ii)    ∵ ∆AMC ≅ ∆BMD
| From (i) above
∠ACM = ∠BDM    | C.P.C.T.
But these are alternate interior angles and they are equal
∴ AC || BD
Now, AC || BD and a transversal BC intersects them
∴ ∠DBC + ∠ACB = 180°
| ∵ The sum of the consecutive interior angles on the same side of a transversal is
180°
⇒ ∠DBC + 90° = 180°
| ∵ ∠ACB = 90° (given)
⇒     ∠DBC = 180° - 90° = 90°
⇒ ∠DBC is a right angle.
(iii)    In ∆DBC and ∆ACB,
∠DBC = ∠ACB (each = 90°)
| Proved in (ii) above
BC = CB    | Common
∵ ∆AMC ≅ ∆BMD    | Proved in (i) above
∴ AC = BD    | C.P.C.T.
∴ ∆DBC ≅ ∆ACB.    | SAS Rule
(iv)    ∵ ∆DBC ≅ ∆ACB
| Proved in (iii) above
∴ DC = AB    | C.P.C.T.

      2CM = AB