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

(2x-3y+z)²=[2x+(-3y)+z]²
=(2x)²+(-3y)²+z²+2\(\times\) 2x\(\times\)(-3y)+2\(\times\)(-3y)\(\times\)z+2\(\times\)2x\(\times\)z
= 4x²+9y²+z²-12xy-6yz+4xz.
 

9.6 cm

40°, 80°, 100°, 140°; Trapezium

\(13;21^{ 0 },70^{ 0 },89^{ 0 }\)

\(\frac{1}{7}=0.142857142857 \ldots=0 . \overline{142857}\) and \(\frac{2}{7}=0.28571428571428 \ldots=0 . \overline{285714}\)
Here, the two decimal expansions are non-terminating recurring.
Hence, 1/7​ and 2/7​ are two rational numbers.
We know, between any two rational numbers, there are infinitely many irrational numbers.
An irrational number has non-terminating non-recurring decimal expansions.
Then an irrational number between \(\frac{1}{7} \text { and } \frac{2}{7}\) is 0.15015001500015.
Similarly, 0.21020020002... is another irrational number between \(\frac{1}{7} \text { and } \frac{2}{7}\)

In Δs APD and CQB
AD = BC (Opp. sides of a parallelogram)
PD = BQ (Given)
∠ADP=∠QBC
⇒ ΔAPD≡ΔCQB
⇒ AP = CQ (c.p.c.t)
In Δs AQB and CPD
AB = DC, BQ = DP
and ∠AQB=∠PDC
ஃ ΔAQB≡ΔCPB ⇒ AQ=CP
(iii) In quad. APCQ, 
AP = CQ and AQ = CP
⇒ APCQ is a parallelogram.

ar(parallelogram ABCD) = AB x AE
= 16 x 8 cm²
= 128 cm²             ...(1)
ar(parallelogram ABCD) = AD x CF
= AD x 10 cm²   .......(2)
From (1) and (2), we get
AD x 10 = 128
AD = \(\frac { 128 }{ 10 } \)
AD = 12.8 cm.

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively.
To Prove:
\((i)\Delta APB\cong \Delta CQD\)
\(\\ (ii)AP=CQ\)
Proof: (i) In \(\Delta APB\ \) and \( \Delta CQD\)
AB = CD                I Opp. sides of || gm ABCD
\(\angle ABP=\angle CDQ\)           |AB || DC and transversal BD intersects them
 \(\angle ABP=\angle CDQ\)           |Each=\(90°\)
(ii)  \(\Delta APB\cong \Delta CQD\)                I Proved above in (i)
 \(\therefore \)  AP = CQ.                I C.P.C.T.

Given: Sides AB and AC of ΔABC are extended to points P and Q respectively.Also, ㄥPBC < ㄥQCB.
To Prove: AC > AB.
Proof: ㄥPBC < ㄥQCB
- ㄥPBC > -ㄥQCB
180⁰ - ㄥ PBC > 180⁰- ㄥQCB
ㄥABC > ㄥACB
AC > AB.

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 \(\angle A\)

and \(\because AB\parallel CD\)
\( CD\parallel EF\) 
\(\because AB\parallel EF\)
Lines parallel to the same line are parallel to each other
\(\therefore x=z\)
Alternate Interior Angles
X+y=\(180^{ 0 }\)
Consecutive interior angles on the same side of a transversal GH to parallel lines AB and CD
From (1) and (2) 
z+y=\(180^{ 0 }\)
y:z=3:7
Sum of the ratios=3+7=10
\(\therefore y=\frac { 3 }{ 10 } X180^{ 0 }=54^{ 0 }\)
and \(z=\frac { 7 }{ 10 } x180^{ 0 }=126^{ 0 }\) 
\(\therefore x=z=126^{ 0 }\)

We have
AC = BD
\(\Rightarrow \) AC - BC= BD - BC
If equals are subtracted from equals, the remainders are equal (Euclid's Axiom (iii))
\(\Rightarrow \) AB = CD
AC - BC coincides with AB; BD - BC coincides with CD [Things which coincide with one another are equal to one another (Euclid's Axiom (iv))]

\(x^3+13x^2+32x+20\)
Let \(p(x)={ x }^{ 3 }+13{ x }^{ 2 }+32x+20\)
By trail, we find that
\(p(-1)={ (-1) }^{ 3 }+13{ (-1) }^{ 2 }+32(-1)+20\)
\(=-1+13-32+20=0\)
\(\therefore\) By Factor Theorem, x-(-1), i.e., (x+1) is a factor of p(x).
Now,
\({ x }^{ 3 }+13{ x }^{ 2 }+32x+20={ x }^{ 2 }(x+1)+12x(x+1)+20(x+1)\)
\(=(x+1)({ x }^{ 2 }+12x+20)\)
\(=(x+1)({ x }^{ 2 }+2x+10x+20)\)
\(=(x+1)\{ x(x+2)+10(x+2)\} \)
\(=(x+1)(x+2)(x+10).\)

Yes! We can predict the decimal expansions of \(\frac { 2 }{ 7 } ,\frac { 3 }{ 7 } ,\frac { 4 }{ 7 } ,\frac { 5 }{ 7 } ,\frac { 6 }{ 7 } \) without actually doing the long division as follows:
To predict the decimal expansion of 2/7, locate when the remainder becomes 2 and respective quotient.Then write the new quotient beginning from there using the repeating digits 1,4,2,8,5,7.
\(\frac { 1 }{ 7 } =0.\overline { 142857 } \)
Similarly,
\(\frac { 2 }{ 7 } =0.\overline { 285714 }\)
\( \\ \frac { 3 }{ 7 } =0.\overline { 428571 }\)
\( \\ \frac { 4 }{ 7 } =0.\overline { 571428 } \)
\(\\ \frac { 5 }{ 7 } =0.\overline { 714285 } \)
\(\\ \frac { 6 }{ 7 } =0.\overline { 857142 } \)

In a trapezium, only one pair of opposite sides is parallel.

Convention

Proof: In \(\triangle\)ABC, BC > AB (.: AB smallest side)
\(\angle\)BAC > \(\angle\)BCA ...(i)
In ACD, CD > AD
(\(\therefore\) CD greatest side) 
\(\angle\)CAD > \(\angle\)ACD ...(ii)
Adding (i) and (ii), we get 
\(\angle\)BAC + \(\angle\)CAD > \(\angle\)BCA + \(\angle\)ACD
\(\Rightarrow\) \(\angle\)BAD > \(\angle\)BCA
\(\angle\)A> \(\angle\)C 
In \(\triangle\)ABD, AD > AB
\(\therefore\) \(\angle\)ABD > \(\angle\)ADB ...(iii) 
In MC;D, CD > BC
\(\therefore\) \(\angle\)DBC > \(\angle\)BDC ...(iv) 
Adding (iii) and (iv), we get
\(\angle\)ABD + \(\angle\)DBC > \(\angle\)ADB +\(\angle\)BDC
\(\Rightarrow\) \(\angle\)ABC > \(\angle\)ADC
\(\Rightarrow\) \(\angle\)B > \(\angle\)D
Hence, \(\angle\)A > \(\angle\)C and \(\angle\)B > \(\angle\)D.

\(p(x)={ \left( \frac { 1 }{ 4 } x \right) }^{ 3 }-{ (2y) }^{ 3 }+\frac { 3 }{ 4 } xy\left[ \frac { 1 }{ 4 } x-2y \right] \)
\(=\left( \frac { 1 }{ 4 } x-2y \right) \left[ { \left( \frac { 1 }{ 4 } x \right) }^{ 2 }+{ (2y) }^{ 2 }+\frac { 1 }{ 4 } x\times 2y \right] +\frac { 3 }{ 4 } xy\left[ \frac { 1 }{ 4 } x-2y \right] \)
\(=\left( \frac { 1 }{ 4 } x-2y \right) \left( \frac { 1 }{ 16 } { x }^{ 2 }+4{ y }^{ 2 }+\frac { 1 }{ 2 } xy+\frac { 3 }{ 4 } xy \right) \)
\(=\left( \frac { 1 }{ 4 } x-2y \right) \left( \frac { { x }^{ 2 } }{ 16 } +{ 4y }^{ 2 }+\frac { 5 }{ 4 } xy \right) \)
\(=\left( \frac { x }{ 4 } -2y \right) \left( \frac { { x }^{ 2 } }{ 16 } +\frac { 1 }{ 4 } xy+xy+4y^{ 2 } \right) \)
\(=\left( \frac { x }{ 4 } -2y \right) \left[ \frac { x }{ 4 } \left( \frac { x }{ 4 } +y \right) +4y\left( \frac { x }{ 4 } +y \right) \right] \)
\(=\left( \frac { x }{ 4 } -2y \right) \left( \frac { x }{ 4 } +y \right) \left( \frac { x }{ 4 } +4y \right) \) 

Two triangles ABC and DEF, such that

AB = DE and AB II DE
Also BC = EF and BC II EF
Proof: (i) In a quadrilateral ABED,
AB = DE and AB II DE
⇒ One pair of opposite sides are equal and parallel.
ABED is a parallelogram
⇒ AD = BE and AD II BE....(i)
(ii) In quadrilateral BCFE,
BC = EF and BC II EF
⇒ One pair of opposite sides are equal and parallel.
BCFE is a parallelogram
CF = BE and CF II BE ...(ii)
(iii) From equations (i) and (ii), we get
AD = CF and AD II CF
⇒ ACFD is a parallelogram
AC = DF andAC II DF
(iv) In ∆ABC and ∆DEF
AB = DE (Given)
BC = EF (Given)
and AC = DF (Proved above in part (c))
So by S.S.s.,ΔABC≅ΔDEF.

\(x=\frac { \sqrt { 5 } +1 }{ \sqrt { 5 } -1 } \)
\({ x }^{ 2 }={ \left[ \frac { \sqrt { 5 } +1 }{ \sqrt { 5 } -1 } \right] }^{ 2 }=\frac { 6+2\sqrt { 5 } }{ 6-2\sqrt { 5 } } =\frac { 3+\sqrt { 5 } }{ 3-\sqrt { 5 } } \)
\(y=\frac { \sqrt { 5 } -1 }{ \sqrt { 5 } +1 } \)
\({ y }^{ 2 }={ \left[ \frac { \sqrt { 5 } -1 }{ \sqrt { 5 } +1 } \right] }^{ 2 }=\frac { 6-2\sqrt { 5 } }{ 6+2\sqrt { 5 } } =\frac { 3-\sqrt { 5 } }{ 3+\sqrt { 5 } } \)
\({ x }^{ 2 }+{ y }^{ 2 }=\frac { { \left( 3+\sqrt { 5 } \right) }^{ 2 }+{ \left( 3-\sqrt { 5 } \right) }^{ 2 } }{ \left( 3-\sqrt { 5 } \right) \left( 3+\sqrt { 5 } \right) } \)
\(=\frac { 9+5+6\sqrt { 5 } +9+5-6\sqrt { 5 } }{ 9-5 } \)
\(=\frac { 28 }{ 4 } \)
\({ x }^{ 2 }+{ y }^{ 2 }=7\)

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.

In \(\triangle \) PRT 
\(\angle PTR\angle PRT+\angle RPT=180^{ 0 }\)
The sum of all angles of a triangle is \(180^{ 0 }\)
\(\Rightarrow \angle PTR+40^{ 0 }+95^{ 0 }=180^{ 0 }\)
\(\Rightarrow \angle PTR+135^{ 0 }=180^{ 0 }\)
\(\Rightarrow \angle PTR+45^{ 0 }\)
\(\Rightarrow \angle QTS=\angle PTR=45^{ 0 }\)
|Vertically Opposite Angles
In TSQ
\(\Rightarrow \angle QTS+\angle TSQ+\angle SQT=180^{ 0 }\)
The sum of all the angles of a triangle is \(180^{ 0 }\)
\(\Rightarrow 45^{ 0 }+75^{ 0 }+\angle SQT=180^{ 0 }\)
\(\Rightarrow 120^{ 0 }+\angle SQT=180^{ 0 }\)
=\(60^{ 0 }\) 

(i) The coordinates of the points B and F are (-5,-4) and (6,0) respectively.
(ii) The abscissae of points D and H are 1 and 0 respectively.
(iv) The ordinates of the points A and C are 1 and 0 respectively.
(iv) The perpendicular distance of the point G from the x-axis is 4 units.

L.H.S=\((a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)\)
\(=\left\{ (a+b)+(b+c)+(c+a) \right\} [(a+b)^2+(b+c)^2+(c+a)^2-(a+b)(b+c-(b_c)(c+a)-(c+a)(a+b)]\)
\(=2(a+b+c)[a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+c^2-ab-ac-b^2-bc-bc-ba-c^2-ca-ca-cb-a^2-ab]\)
Using Identity I
\(=2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\)
\(=2(a^3+b^3+c^3-3abc)\)Using Identity VIII