(a) 60⁰
(b) 60⁰

(i) Number of triangles = 25 x 6
= 25 + 25 + 25 + 25 + 25 + 25 = 150

(ii) Number of triangles = 25 x 6 + 25 x 6
= 150 + 150
= 300

Figure (ii) has more triangles.

Given: In figure,
ㄥB < ㄥA and ㄥC < ㄥD.
To Prove: AD < BC
Proof: ㄥB < ㄥA
ㄥA > ㄥB
OB> OA ...(1)
ㄥC < ㄥD
ㄥD > ㄥC
OC > OD  ..(2)
| Side opposite to greater angle is longer
From (1) and 2), we get
OB+OC > OA+OD
⇒ BC> AD
⇒ AD < Be.

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

(i) In ABC, we have
AB = AC [Given]
∴ ∠C = ∠B
[Angle opposite to equal sides are equal]
⇒ \(\frac{1}{2}\)∠C = \(\frac{1}{2}\)∠B
or ∠OCB = ∠OBC
⇒ OB = OC 
[Sides opposite to equal angles are equal]
(ii) In ΔABO and ΔACO, we have
AB =AC [Given]
OB = OC [Proved
∠OBA = ∠OCA  [∵ \(\frac{1}{2}\)∠B-\(\frac{1}{2}\)∠C]
∴ Using SAS criteria,
ΔABO =ΔACO
⇒ ∠OAB = ∠OAC [c.p.c.t.]
⇒ AO bisects LA. 

AD = BD
\(\Rightarrow\) \(\angle\)ABD = \(\angle\)DAB = 59⁰
(Angles opp. to equal sides are equal) 
In \(\triangle\)ABD,
59⁰ + 59⁰ + \(\angle\)ADB = 180⁰
\(\Rightarrow\) \(\angle\)ADB = 180⁰- 118⁰ = 62⁰
and \(\angle\)ACD = 62⁰- 32⁰ = 30⁰
(Exterior angle is equal to the sum of interior opposite angles)
In \(\triangle\)ABD,
(Side opp. to greatest angle is the longest) 1
Also in \(\triangle\)ABC, AB < AC
BD < AC

Let, BE\(\bot \)AC and CF\(\bot \)AB
In \(\triangle\)s ABE and ACF, we have
\(\angle\)AEB = \(\angle\)AFC (.: Each = 90°) 
\(\angle\)A = \(\angle\)A (Common)
and BE = CF (Given)
\(\therefore\) By AAS criterion at congruence, we have
\(\triangle ABE\cong \triangle ACF\)
\(\Rightarrow\) AB = AC.
(\(\therefore\) Corresponding parts of congruent triangles are equal)

Let AB be the height of Rahim's building. The length of his bamboo stair is AC. The carpenter places the foot of the stair at C at a distance of BC from the wall.
Then, MBC is a right angled triangle.
Also, let height of Me era's building be PQ, which is equal to AB, if she may keep the foot R of her stair at a same distance, i.e. at QR = BC
Also \(\angle ABC=\angle PQR=90^{ 0 }\)
AB = PQ
\(\triangle ABC\equiv \triangle PQR\)
[by SAS congruence rule]
Hence, she use SAS congruence rule to make her bamboo stair without taking measurement of the bamboo stair
Since \(\triangle ABC\equiv \triangle PQR\)
AC = PR [by CPCT]
Thus, length of the stairs are equal. Here, values depicted are friendly and cooperative nature between two friends ignoring their religion.

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

In \(\triangle OAP\) and \(\triangle OBQ\)
AP = BQ
\(\angle OAP=\angle OBQ\)
\(\angle AOB=\angle BOQ\) | Vertically opposite angles
\(\triangle OAP\cong \triangle OBQ\) | AAS Rule
OA = OB | C.P.C.T
OP = OQ | C.P.C.T
O is the mid-point of line segments AB and PQ.

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

Since, AD = AE
∴ ㄥADE=ㄥAD
[ ∵ angles opposite to equal sides of a triangle are equal]
Now, 180⁰ - ㄥADE = 180⁰-ㄥAED
⇒ ㄥADB =ㄥAEC  ..(i)
Now, prove that ΔADB\(\cong \)ΔAEC with the help of Eq. (i) by SAS congruence rule, then show that AB = AC by CPCT.

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

Theorem