\(\angle\)A,  \(\angle\)AEB = \(\angle\)AFC = 90⁰
\(\therefore\) \(\angle\)ABE = \(\angle\)ACF, BE = CF
\(\therefore\) \(\triangle AEB\cong \triangle AFC\) (ASA) 
\(\therefore\) AB = AC
Alternative method:
In \(\triangle\)BCE and \(\triangle\)CBF,
\(\angle\)BEC = \(\angle\)BFC = 90⁰    (Given)
BE = CF   (Given)
BC = CB   (Common)
\(\therefore \triangle BCE\cong \triangle CBF\)
\(\angle\)B =\(\angle\)C
AC =AB   (By c.p.c.t)
Similarly, \(\triangle ABD\cong \triangle BAE\)
AC = BC
Therefore, AB = BC = AC
Thus, \(\triangle\)ABC is an equilateral triangle.

Given: ΔABD and ΔBCD and isosceles
triangles on the same base BD.
To Prove: ΔABC = ΔADC
Proof: ΔABD is isosceles
AB = AD
ΔABD =  ΔADB ... (1)
|Angles opposite to equal sides of a triangle are equal
ΔCBD is isosceles
CB = CD
ΔCBD = ΔCDB ... (2)
|Angles opposite to equal sides of a triangle are equal
Adding (1) and (2), we get,
ΔABD + ΔCBD = ΔADB +Δ CDB
⇒ ΔABC = ΔADC

Given: In ΔABC, D is the mid-point of BC The perpendiculars from D to AB and AC are equal.
To Prove: ΔABC is isosceles.

Proof: In righ~ triangles DEC and DFB, Hyp. DC = Hyp. DB
| D is the mid-point of BC
Side DE = Side DF     IGiven
∴ ΔDEC ≌ ΔDFB I RHS congruence rule
∴ ㄥDCE = ㄥDBF
∴ ㄥBCA = ㄥCBA
∴ AB = AC
|Sides opposite to equal angles of a triangle are equal

In \(\triangle OAC\) and \(\triangle ODB\)
OA = OD
OB = OC
\(\angle AOC=\angle DOB\) | Vertically opposite angles

\(\triangle OAC\cong \triangle ODB\)
AC = BD
\(\angle OAC=\angle ODB\)
\(\angle OCA=\angle OBD\)
thus \(\angle OAC\) may not be equal to \(\angle OBD\) and therefore, AC may not be parallel to BD
However, if OA = OC, then
\(\angle OAC=\angle OCA\)
\(\angle OAC=\angle ODB\)
But abgles from a pair of equal alternate angles \(AC\parallel BD\)

Given: AE bisects \(\angle DAC\) and,\(\angle B=\angle C\)
To prove: \(AE\parallel BC\)
Proof: In \(\angle ABC\)
Ext.\(\angle DAC=\angle ABC+\angle ACB\) | An exterior angle of a triangle is equal to the sum of its two interior opposite angles
\(\angle DAC=\angle ACB+\angle ACB\)
\(\angle DAC=2\angle ACB\)
\(2\angle CAE=2\angle ACB\)
\(\angle CAE=\angle ACB\)
But these angles from a pair equal alternate interior angles
\(AE\parallel BC\)

Given: A triangle ABC in which AB = AC
To Prove: \(\angle ABC=\angle ACB\)

Construction: Draw the bisector AD of A so as to intersect BC at D.
Proof: In \(\triangle ADB\) and \(\triangle ADC\)
AD = AC
\(\angle BAD=\angle CAD\)
\(\triangle ADB\cong \triangle ADC\) | SAS congruence rule
\(\angle ABD=\angle ACD\) | C.P.C.T
\(\angle ABC=\angle ACB\)
Yes, the converse is true

Given:  \(\angle QPR=\angle PQR\) and M and N are respectively points on sides QR and PR of  \(\triangle PQR\) , such that QM = PN.
To prove: OP = OQ, where O is the point of intersecting of PM and QN.
Proof: In \(\triangle PNQ\) and \(\triangle QMP\)
PN = QM
PQ = QP
\(\angle QPN=\angle PQM\)
\(\triangle PNQ=\triangle QMP\) | SAS congruence rule
\(\angle PNQ=\angle QMB\) | C.P.C.T
Again, in \(\triangle PNO\) and \(\triangle QMO\)
PN = QM
\(\angle PON=\angle QOM\)
\(\angle PNO=\angle QMO\) | proved above
\(\triangle PNO\cong \triangle QMO\) | AAS congruence rule
OP = OQ | C.P.C.T

Given:
AB = FE
BC = ED
\(AB\bot BD\)
\(FE\bot EC\)
To prove: (i) \(\triangle ABD\cong \triangle FEC\)
(ii) \(AD\cong FC\)
Proof: (i) In \(\triangle ABD\) and \(\triangle FEC\)
AB = FE ......... (1)
\(\angle ABD=\angle FEC\) ... (2)   | Each 90⁰
BC = ED
BC + CD = ED + DC
BD = EC ....... (3)
In view of (1), (2) and (3)
\(\triangle ABD\cong \triangle FEC\)  | SAS congruence rule
(ii) AD = FC | C.P.C.T

Given: In a rectangle ABCD, E is a point which bisects BC
To Prove: AE = ED

Proof: In \(\triangle EBA\) and \(\triangle ECD\)
EB = EC
\(\angle EBA=\angle ECD\) | Each 90⁰ (ABCD is a rectangle)
BA = CD | Opposite sides of rectangle ABCD
\(\triangle EBA\cong \triangle ECD\) | SAS congruenece rule
AE = DE | C.P.C.T
AE = ED