In \(\angle\)ABD and \(\angle\)ACD,
\(\angle\) BAD = \(\angle\)CAD (Given)
AD = AD (Common)
\(\angle\) ADB = \(\angle\)ADC = 90° (Given)
So, \(\angle\) ABD  \(\angle\) ACD (ASA rule)
So, AB = AC (CPCT)
or, \(\angle\) ABC is an isosceles triangle.

We have, P is the mid-point of AB.
∴  AP = BP
∠EPA = ∠DPB [Given]
Adding ∠EPD on both sides, we get
∠EPA + ∠EPD = ∠DPB + ∠EPD
⇒  APD = ∠BPE
(i) Now, in ΔDAP ≅ ΔEBP, we have
AP= BP [Proved]
∠PAD = ∠PBE [.: It is given that LBAD = LABE]
∠DPA = ∠EPB [Proved]
∴ Using ASA criteria, we have
ΔDAP ≅ ΔEBP
(ii) Since, ΔDAP ≅ ΔEBP
∴ Their corresponding parts are equal.
⇒ AD = BE.

We have ∠BAD = ∠EAC
Adding ∠DAC on both sides, we have
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠DAE
Now, in ΔABC and ΔADE, we have
∠BAC = ∠DAE [Proved]
AB = AD [Given]
AC = AE [Given]
∴ ΔBC ≅ ΔADE [Using SAS criteria]
SinceΔABC ≅ ΔADE, therefore, their corresponding parts are equal.
⇒ BC = DE.

Proof: In \(\triangle\)ABF and \(\triangle\)ACE,
AB =AC
\(\angle\)A=\(\angle\)A
AF =AE
\(\therefore\) \(\triangle\)ABF = \(\triangle\)ACE (by SAS cong.)
\(\therefore\) By c.p.c.t BF = CE.
Alternative Method:
AB = AC\(\Rightarrow \frac { AB }{ 2 } =\frac { AC }{ 2 } \)
\(\Rightarrow \) AE=AF,
sice E and F are the mid-points of AB and AC
In \(\triangle\)ABF and \(\triangle\)ACE,
AB = AC (Given) 
\(\angle\)A = \(\angle\)A (Common) 
AF = AE (Proved) 
\(\therefore \triangle ABF\cong \triangle ACE\) (By SAS cong.) 
\(\therefore\) BF = CE. (By c.p.c.t) 

In \(\triangle\)ABE and \(\triangle\)ACD
AB =AC    (Given)
\(\angle\)C = \(\angle\)B
(Angle opp. to equal sides) 
BE = CD (Given)
\(\therefore \ \triangle ABE\cong \triangle ACD\) (By SAS) 
AO = AE. (By c.p.c.t.)

AB =AC
\(\therefore\) \(\angle\)B = \(\angle\)C = x
BA =BC
\(\therefore\) \(\angle\)A = \(\angle\)C = x
AC =BC
\(\angle\)A = \(\angle\)B = x
But, \(\angle\)A + \(\angle\)B + \(\angle\)C = 180°
Also,
3x = 1800
\(\angle\)A = \(\angle\)B = \(\angle\)C = 60°.
Alternative Method:

Let \(\triangle\)ABC be an equilateral triangle, so that AB = AC = BC
Now, AB = AC
\(\Rightarrow\) \(\angle\)B = \(\angle\)C ....(1)
(\(\because\)Angles opp. to equal sides are equal)
CB = CA
\(\Rightarrow\) \(\angle\)A = \(\angle\)B ....(2)
(\(\because\) Angles opp. to equal sides are equal)
From (1) and (2), we have
\(\angle\)A = \(\angle\)B = \(\angle\)C
Also, \(\angle\)A +\(\angle\).B +\(\angle\)C = 180°,(Angle sum property)
\(\therefore\) \(\angle\)A + \(\angle\)A + \(\angle\)A = 180°
\(\Rightarrow\) 3\(\angle\)A = 180° \(\Rightarrow\) \(\angle\)A = 60°.
\(\therefore\) \(\angle\)A = \(\angle\)B = \(\angle\)C = 60°.
Thus, each angle of an equilateral triangle is 60°.

In \(\triangle\)PAM and \(\triangle\)QBM,
As M is the mid-point,
\(\Rightarrow\) AM = BM
\(\angle\)1 = \(\angle\)2
\(\angle\)3 = \(\angle\)4 = 90°
\(\therefore\) \(\triangle PAM\cong \triangle QBM\)  (By ASA Cong.) 
\(\therefore\) PA = BQ. (By c.p.c.t.) Proved.

\(\angle\)DOC = \(\angle\)AOB 
Proving \(\triangle COD\cong \triangle BOA\) (SAS) 
CD = AB. (c.p.c.t.)
Alternative Method:
In \(\triangle\)COD and \(\triangle\)BOA,
OD =OA (Given)
OC =OB (Given)
\(\angle\)DOA = \(\angle\)COB = 90°
\(\therefore\) \(\angle\)DOA + \(\angle\)AOC = \(\angle\)COB + \(\angle\)AOC
i.e., \(\angle\)DOC = \(\angle\)AOB
\(\therefore\) \(\triangle COD\cong \triangle BOA\)     (By SAS)
\(\Rightarrow\) CD =AB.  (By c.p.c.t.)

\(\angle\)BDC + \(\angle\)CDA = 180°   (Linear Pair)
\(\angle\)BDC + x = 180°
\(\angle\)BDC = 180° - x 
Similarly, \(\angle\)BEA = 180° - y
x = Y (Given)
\(\angle\)BDC = \(\angle\)BEA 
\(\angle\)B = \(\angle\)B (Common)
AB = BC
\(\triangle BAE\cong \triangle BCD\) (ASA) 
\(\therefore\)  AE = CD (c.p.c.t.)

PQRS is a square. (given)

(i) SRT is an equilateral triangle. (given)
\(\therefore\) \(\angle\)PSR = 90°, \(\angle\)TSR = 60° 
\(\Rightarrow\) \(\angle\)PSR + \(\angle\)TSR = 150°.
Similarly, \(\angle\)QRT = 150°
In \(\triangle\)PST and \(\triangle\)QRT, we have PS = QR (S)
\(\angle\)PST =\(\angle\)QRT = 150° (A)
and ST = RT (5) Y,
By SAS, \(\triangle PST\cong \triangle QRT\)
\(\Rightarrow\) PT = QT (By c.p.c.t.) Proved.
(ii) In \(\triangle\)TQR, QR = RT
(Square and equilateral 6 on same base) Yz
\(\Rightarrow\) \(\angle\)TQR = \(\angle\)QTR = x 
\(\therefore\) x + x + \(\angle\)QRT = 180°
\(\Rightarrow\) 2x + 150° = 180° ~ 2x = 30°
\(\therefore\) x = 15° Proved.

In \(\triangle\)ABC and \(\triangle\)BAD,
BC =AD   (Given)
\(\angle\)CBA = \(\angle\)BAD  (Given)
AB =AB  (Common)
\(\triangle\)ABC \(\cong \) \(\triangle\)BAD (By SAS cong.)
BD = AC (By c.p.c.t.) 

BC = DE
Adding CD to both sides
BC + CD = DE + CD
\(\Rightarrow\)BD =CE
In \(\triangle\)ABD and \(\triangle\)FEC,
AB = EF (Given)
\(\angle\)B = \(\angle\)E = (90°) 
BD = CE (proved)
\(\triangle ABD\cong \triangle FEC\) (by SAS)

BR + BQ = CQ + BQ \(\Rightarrow\) QR = BC
\(\triangle ABC\cong \triangle PQR\) by RHS\(\cong\)condn
\(\therefore\) AC = PR (By c.p.c.t.)

In \(\triangle\)sABC and DCB,
AB = DC
\(\angle\)ABC = \(\angle\)DCB {(Given)}
BC = CB (Common) 
\(\triangle ABC\cong \triangle DCB\) (SAS) 
AC = DB. (By c.p.c.t.)

In \(\triangle\)ABD and \(\triangle\)CDB,
AB = CD (Given)
\(\angle\)ABD = \(\angle\)CDB (Given) 
BD = BD (Common)
\(\Rightarrow\) \(\triangle ABD\cong \triangle CDB\) (By SAS)
\(\Rightarrow\) AD = CB. (By c.p.c.t.)

40⁰

Let ABC be a right angled triangle in which ㄥB = 90°.
Then, ㄥA +ㄥC = 90°
|Sum of all the angles of a triangle is 180°
ㄥB=ㄥA+ㄥC
ㄥB>ㄥA and ㄥB>ㄥC
AC>BC
| Side opposite to greater angle is longer
and AC> AB

AC is the longest side, i.e., hypotenuse is the longest side.

Given: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \(\angle PQR\)
To prove: (i) \(\triangle ABM\cong \triangle PQN\) (ii) \(\triangle ABC\cong \triangle PQR\)
Proof: In \(\triangle ABM\) and \(\triangle PQN\)
AB = PQ
AM = PN
BC = QR
2BM = 2QN | M and Nare the mid-points of BC and QR respectively
BM = QN
In view of (1), (2) and (3)
\(\triangle ABM\cong \triangle PQN\) | SSS rule
(ii) \(\triangle ABM\cong \triangle PQN\)
\(\angle ABM=\angle PQN\) | C.P.C.T
\(\angle ABC=\angle PQR\)
In \(\triangle ABC\) and \(\triangle PQR\)
AB = PQ
BC = QR
\(\angle ABC=\angle PQR\)
\(\triangle ABC=\triangle PQR\) | SAS rule

Given: AD is an altitude of an isosceles triangle ABC in which AB = AC.
To prove: (i) AD bisects BC (ii) AD bisects \(\angle A\)
Proof: (i) In right \(\triangle ADB\) and right \(\triangle ADC\)
Hyp.AB = Hyp. AC
Side AD = Side AD

\(\triangle ADB\cong \triangle ADC\) | RHS rule
BD = CD | C.P.C.T
AD bisects BC
(ii) \(\triangle ADB\cong \triangle ADC\)
\(\angle BAD=\angle CAD\)
AD bisects \(\angle A\)

Given: An equilateral triangle ABC
To prove: \(\angle A+\angle B+\angle C={ 60 }^{ 0 }\)
Proof: ABC is an equilateral triangle
AB = BC = CA ....... (1) | 
AB = BC
\(\angle A=\angle C\) .......... (2) | Angles opposite to equal sides of a triangle are equal
BC = CA
\(\angle A=\angle B\) ......... (3) | Angles opposite to equal sides of a triangle are equal
From (2) and (3), we obtain
\(\angle A=\angle B=\angle C\)  ........ (4)
In \(\triangle ABC\)
\(\angle A+\angle B+\angle C={ 180 }^{ 0 }\) ...... (5) | Sum of all the angles of a triangle is 180°
Let \(\angle A={ x }^{ 0 }\) then, \(\angle B=\angle C={ x }^{ 0 }\)
From (5)
\({ x }^{ 0 }+{ x }^{ 0 }+{ x }^{ 0 }={ 180 }^{ 0 }\)
\(3{ x }^{ 0 }={ 180 }^{ 0 }\)
\({ x }^{ 0 }={ 60 }^{ 0 }\)
\(\angle A=\angle B=\angle C={ 60 }^{ 0 }\)

Given: \(\triangle ABC\) is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB
To prove:   \(\angle BCD\)  a right angle.
Proof: ABC is an isosceles triangle
\(\angle ABC=\angle ACB\)
AB = AC AND AD = AB
AC = AD
In \(\triangle ACD\)
\(\angle CDA=\angle ACD\) | Angles opposite to equal sides of a triangle are equal
\(\angle CDB=\angle ACD\)
Adding the corresponding sides of (1) and (2) we get
\(\angle ABC+\angle CDB=\angle ACB+\angle ACD\)
\(\angle ABC+\angle CDB=\angle BCD\)
In  \(\triangle BCD\)
\(\angle BCD+\angle ABC+\angle CDB={ 180 }^{ 0 }\)
\(\angle BCD+\angle BCD={ 180 }^{ 0 }\)
\(2\angle BCD={ 180 }^{ 0 }\)
\(\angle BCD={ 90 }^{ 0 }\)
\(\angle BCD\)  is a right angle.

Given: ABC and DBC are two isosceles triangles on the same base BC
To prove: \(\angle ABD=\angle ACD\)
Proof: ABC is an isosceles triangle on the base BC
\(\angle ABC=\angle ACB\)
DBC is an isosceles triangle on the base BC
\(\angle DBC=\angle DCB\)
Adding the corresponding sides of (1) and (2) we get
\(\angle ABC+\angle DBC=\angle ACB+\angle DCB\)
\(\angle ABD=\angle ACD\)

Given: Line l is the bisector of an angle\(\angle A\) and  \(\angle B\)   is any point on I.BP and BQ are perpendiculars from B to the arms of  \(\angle A\)
To Prove: (i) \(\triangle APB\cong \triangle AQB\)
(ii) BP = BQ or B is equidistant from the arms of \(\angle A\)
Proof: (i) In \(\triangle APB\) and \(\triangle AQB\)
\(\angle BAP=\angle BAQ \)  | l is the bisector of \(\angle A\)
AB = AB | Common
\(\angle BAP=\angle BAQ \)  | Each = 90⁰
BP = BQ or B is equidistant from the arms of \(\angle A\)
\(\triangle APB\cong \triangle AQB\)   | SAS Rule
(ii) \(\triangle APB\cong \triangle AQB\) | Proved in (i) above
BP = BQ  | C.P.C.T

Given: l and m are two parallel lines intersected by another pair of parallel lines p and q
To Prove: \(\triangle ABC\cong \triangle CDA\)
Proof: \(AB\parallel DC\) and \(AD\parallel BC\)
Quadrilateral ABCD is a parallelogram.
| A quadrilateral is a parallelogram if both the pairs of opposite sides are parallel
BC = AD ........... (1) | Opposite sides of a \(\parallel \) gm are equal
AB = CD .............(2) | Opposite sides of a \(\parallel \) gm are equal
\( \angle ABC=\angle CDA\)......(3)  | Opposite angles of a \(\parallel \) gm are equal
In  \(\triangle ABC\) and \(\triangle CDA\),
AB = CD | From (2)
BC = DA | From (1)
\( \angle ABC=\angle CDA\)  | From (3)
\(\triangle ABC\cong \triangle CDA\) | SAS Rule

Given ABCD is a quadrilateral in which AD=BC and \(\angle DAB=\angle CBA\) 
To prove: (i) \(\triangle ABD\cong \triangle BAC\)
(ii) BD = AC
(iii) \(\angle ABD=\angle BAC\)
Proof: (i) In \(\triangle ABD\) and \(\triangle BAC\)
AD = BC
AB = BA
\(\angle DAB=\angle CBA\)
\(\triangle ABD\cong \triangle BAC\)  |SAS Rule
(ii) \(\triangle ADB\cong \triangle BAC\)  |Proved in (i)
BD = AC  | C.P.C.T
(iii) \(\triangle ABD\cong \triangle BAC\) |Proved in (i)
\(\angle ABD=\angle BAC\)  | C.P.C.T