In \(\triangle\)ABC, AC >AB
\(\therefore\) \(\angle\) ABC > \(\angle\)ACB
(Angles opposite to larger side is greater)
\(\therefore\) \(\angle\)ABC + \(\angle\)1 > \(\angle\)ACB + \(\angle\)1
(Adding \(\angle 1\)on both sides) Y.
\(\therefore\) \(\angle\)ABC + \(\angle\)1 > \(\angle\)ACB + \(\angle\)2
(AD bisects \(\angle\)A, \(\angle\)1 = \(\angle\)2) 
\(\therefore\) \(\angle\)ADC > \(\angle\)ADB.
(Exterior angle property of triangle)

In \(\triangle\)PQS and \(\triangle\)PRS,
PQ = PR  (Given)
PS = PS   (Common)
\(\angle\)PSQ = \(\angle\)PSR = 90°
(PS is altitude)
By R.H.S. rule,
\(\triangle PQS\cong \triangle PRS\)
\(\angle\)QPS = \(\angle\)RPS (By c.p.c.t.) 
Hence, PS bisects \(\angle\)P.

Given, AX II BY
\(\angle\)BAX = \(\angle\)ABY (Alternate angles)    ...(i)
\(\angle\)AXY = \(\angle\)BYX (Alternate angles)    ... (ii) 
In \(\triangle\)APX and \(\triangle\)BPY,
AX = BY (Given) ...(iii) 
From (1), (2) and (3), we get
\(\triangle APX\cong \triangle BPY\)
(By ASA) Proved

Let, \(\angle\)DAC be \(\angle\)3
\(\angle\)1= \(\angle\)2
\(\angle\)1 + \(\angle\)3 = \(\angle\)2 + \(\angle\)3
\(\angle\)BAC = \(\angle\)EAD    ...(i)
Given that, BD = CE
BD + DC = CE + DC
\(\Rightarrow\) BC = DE      ....(ii)
\(\angle\)B = \(\angle\)E       (Given)....(iii)
From (i), (ii) and (iii), we get
\(\triangle ABC\cong \triangle AED\) (By AAS rule).

(i) In \(\triangle\) AOD and \(\triangle\)BOC,
OA =OB (Given)
OD = OC (Given)
\(\angle\)AOD = \(\angle\)BOC
(Vertically opposite angles)
So, by SAS criteria,
\(\triangle AOD\cong \triangle BOC\)
(ii) \(\angle\)CBA = \(\angle\)DAB (By c.p.c.t.)
AD and BC are two lines intersected by AB such that \(\angle\)CBA = \(\angle\)DAB and they form a pair of alternate angles.
Hence, AD II BC

In \(\triangle\)PAB and \(\triangle\)PDC, 
PA = PD (Given)
(P is the mid-point of AD)
AB = CD (Side of a square)
\(\angle\)PAB = \(\angle\)PDC = 90°
By R.H.S., \(\triangle PAB\cong \triangle PDC\)
\(\therefore\) PB = PC (By c.p.c.t.)
(Angles opp. to equal sides are equal)
\(\Rightarrow\) \(\angle\)PCB = \(\angle\)PBC. Proved.

In \(\triangle\)BED and \(\triangle\)CFD,
\(\angle\)DEB = \(\angle\)DFC = 90° 
BD = DC (D is the mid-point)
ED = FD (Given)
\(\therefore \triangle BED\cong \triangle CFD\) (By RHS) 
\(\Rightarrow\) \(\angle\)B = \(\angle\)C (By c.p.c.t.) Proved.

OA = OB (O is the mid-point of AB)
\(\angle\)AOC = \(\angle\)BOD (Vertically opposite angles)
OC = OD (O is the mid-point of CD)
\(\triangle AOC\cong \triangle BOD\)
\(\Rightarrow\) AC = BD. (By c.p.c.t) Proved.

In \(\triangle\)ADC and \(\triangle\)ABC,
AC is Common
\(\angle DAC=\angle BAC\)
\(\angle DCA=\angle BCA\)(Given)
Hence, \(\triangle ADC\cong \triangle ABC\) (By AAS rule)
\(\Rightarrow\) CD = BC (By c.p.ct) Proved

Join AD.
In \(\triangle\)ABC and \(\triangle\)ACD,
AB = AC   (Given)
BD = CD   (Given)
AD = AD  (Common)
By using SSS Congruency Rule,
\(​​\triangle ABD\cong \triangle ACD\)
\(\therefore \angle ABD=\angle ACD (By c.p.c.t)\)

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