(a) 60⁰
(b) 60⁰

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 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

QR as ㄥP is the greatest

AC as ㄥB is the smallest.

DE, EF

BC, AB

Given: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively.
To Prove: BE = CF
Proof: ABC is an isosceles triangle
AB = AC
\(\angle ABC=\angle ACB\) | Angles opposite to equal sides of a triangle are equal
In \(\triangle BEC\) and \(\triangle CFB\)
\(\angle BEC=\angle CFB\)  | Each = 90⁰
BC = CB
\(\angle ECB=\angle FBC\)
\(\triangle BEC\cong \triangle CFB\) | By AAS Rule
BE = CF | C.P.C.T

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

RT = TS
\(\angle TRS=\angle TSR\) ....... (1) | Angles opposite to equal sides of a triangle are equal
\(\angle 1=\angle 4\) | vertically opposite angles
\(2\angle 2=2\angle 3\)
\(\angle 2=\angle 3\) ......... (2)
subtracting (2) from (1), we get
\(\angle TRS-\angle 2=\angle TSR-\angle 3\)
\(\angle TRB=\angle TSA\)
RT = ST
\(\triangle RBT\cong \triangle SAT\)  | AAS congruence rule
(ii) RB = SA  | C.P.C.T
RB = AS

ㄥBAC=180⁰-(35⁰+65⁰)=80⁰
ㄥBAX=40⁰
ㄥBAX > ㄥABX
BX > AX

BC > AC > AB
ㄥA > ㄥB
⇒ ㄥB < ㄥA

Theorem

Obviously(b)

Two circles of the same radii are congruent

≌ represents congruence.