Solution Medians are line segments joining each vertex to the midpoint of the corresponding opposite sides.
Thus medians are the cevians where D, E, F are midpoints of BC, CA and AB respectively
Since D is a midpoint of BC, BD = DC so \(\frac { BD }{ DC } =1\)        ..(1)
Since, E is a midpoint of CA,CE = EA so \(\frac { CE }{ EA } =1\)       ..(2)
Since, F is a midpoint of AB, AF FB so \(\frac { AF }{ FB } =1\)       ...(3)
Thus, multiplying (1), (2) and (3) we get,
\(\frac { BD }{ DC } \times \frac { CE }{ EA } \times \frac { AF }{ FB } =1\times 1\times 1=1\)
And so, Ceva’s theorem is satisfied.
Hence the Medians are concurrent.

\(\therefore\) AB touches the smaller circle at P
\(\therefore \mathrm{OP} \perp \mathrm{AB} \Rightarrow \angle O P A=90^{\circ}\)
Now AB is the chord of the bigger circle. Since the perpendicular from the centre to a chord, bisects the chord.
\(\therefore\) P is the mid point of AB
\(\mathrm{AP}=\frac{8}{2}=4 \mathrm{~cm}\)
\(\text { In right } \triangle A P O \text {, we have }\)
\(\mathrm{AO}^{2}=\mathrm{OP}^{2}+\mathrm{AP}^{2} \)
\(\mathrm{AO}^{2}=3^{2}+4^{2}=9+16 \)
\(\mathrm{AO}^{2}=25 \)
\(\mathrm{AO}=5 \mathrm{~cm} \)
\(\therefore \text { The radius of the bigger circle }=5 \mathrm{~cm} \text {. }\)

Let NM be the chord of a circle with centre C. Let the tangents at M and N meet at O
\(\therefore\) OM is a tangent at M
\(\therefore\)  \(\angle\)OMC = 90^o
Similarly \(\angle\)ONC : 90^o
Since CM = CN
In \(\Delta\)CMN \(\angle\)1 = \(\angle\)2
[\(\because\) Angles opposite to equal sides are equal in a triangle]
\(\angle\)OMC - \(\angle\) 1 = \(\angle\)ONC - \(\angle\)2
\(\angle\)OML = \(\angle\)ONL
Thus tangents make equal angles with the chord.

Let ABC be an equilateral triangle and AD \(\perp\) BC.
In \(\Delta\)ADB and \(\Delta\)ADC , we have
AB = AC
\(\angle\)B = \(\angle\)C
and \(\angle\)ADB = \(\angle\)ADC
By RHS Criteria
\(\triangle A D B \cong \triangle A D C\)
\(\mathrm{BD}=\mathrm{DC} \ [\because \text { By CPCTC] }\)
\(\Rightarrow \mathrm{BD}=\mathrm{DC}=\frac{1}{2} \mathrm{BC}\)
\(\text { Since } \triangle A D B \text { is right triangle right angled at } \mathrm{D}\)
\(A B^{2} =A D^{2}+B D^{2} \)
\(\Rightarrow A B^{2} =A D^{2}+\left(\frac{1}{2} B C\right)^{2} \)
\(\Rightarrow A B^{2} =A D^{2}+\frac{B C^{2}}{4} \)
\(A B^{2} =A D^{2}+\frac{A B^{2}}{4} \quad[\because B C=A B] \)
\(\mathrm{AD}^{2} =\mathrm{AB}^{2}-\frac{A B^{2}}{4} \)
\(=\frac{4 A B^{2}-A B^{2}}{4} \)
\(\mathrm{AD}^{2} =\frac{3 A B^{2}}{4} \)
\(4 \mathrm{AD}^{2} =3 \mathrm{AB}^{2} \)

Let AB be the width of the street; C be the foot of the ladder. Let D and E be the windows at heights of 9 m and 12 m respectively from the ground.
Then CD and CE are the two positions of the ladder.
Clearly AD = 9m,BE = 12m,CD = CE = 15m
From the right triangle \(\Delta\)ACD we have
\(C D^{2} =A C^{2}+A D^{2} \)
\(15^{2} =A C^{2}+9^{2} \)
\(A C^{2} =225-81=144 \)
\(A C =12 \mathrm{~m} \)
\(\text { In } \triangle B C E \text {, we have }\)
\(C E^{2} =B C^{2}+B E^{2} \)
\(15^{2} =B C^{2}+12^{2} \)
\(B C^{2} =225-144=81 \)
\(B C =9 \mathrm{~m} \)
\(\text { Hence width of the street } \mathrm{AB}=\mathrm{AC}+\mathrm{CB}\)
\(=12+9=21 \mathrm{~m}\)

Given: A quadrilateral ABCD in which the diagonal BD bisects \(\angle\)B and \(\angle\)D
\(\text { To prove: } \frac{A B}{B C}=\frac{A D}{C D}\)
Construction: join AC intersecting BD in O.
Proof
In \(\Delta\)ABC , BO is the bisector of \(\angle\)B.
\(\frac{A O}{O C}=\frac{B A}{B C} \)
\(\frac{O A}{O C}=\frac{A B}{B C} \)
\(\text { In } \triangle A D C, \mathrm{DO} \text { is the bisector of } \angle D\)
\(\frac{A O}{O C}=\frac{D A}{D C} \)
\(\frac{O A}{O C}=\frac{A D}{C D} \)
\(\text { From (1) and (2) we get }\)
\(\frac{A B}{B C}=\frac{A D}{C D}\)

Given: In \(\Delta\)ABC AD and AE respectively the bisectors of the interior and exterior angles at A.
\(\text { To prove: } \frac{B D}{B E}=\frac{C D}{C E}\)
Proof : Since AD is the internal bisector of \(\angle\)A meeting BC at D.
\(\frac{A B}{A C}=\frac{B D}{D C}\)
Since AE is the external bisector of \(\angle\)A meeting BC produced in E.
\(\frac{A B}{A C}=\frac{B E}{C E}\)
From (1) and (2) we get
\(\frac{B D}{D C}=\frac{B E}{C E} \)
\(\frac{B D}{B E}=\frac{C D}{C E} \)

We have an isosceles triangle \(\Delta\)ABC in which
AB = AC.
\(\text { In } \triangle A B D \text { and } \Delta E C F\)
\(\mathrm{AB}=\mathrm{AC}\)
\(\Rightarrow \text { Angles opposite to them are equal. }\)
\(\therefore \angle A C B =\angle A B C \)
\(\angle E C F =\angle A B D\)
\(\mathrm{AD} \perp \mathrm{BC} \text { and } \mathrm{EF} \perp \mathrm{AC}\)
\(\angle A D B=\angle E F C=90^{\circ}\)
From (1) and (2) we have
By AA criteria of similarity
\(\triangle A B D \sim \triangle E C F\)

T is a point on QR and S is a point or PR such that
\(\angle R T S=\angle P\)
\(\text { Now in } \triangle R P Q \text { and } \triangle R T S\)
\(\angle R P Q=\angle R T S \)
\(\angle P R Q=\angle T R S \)
\(\therefore \text { Using AA similarity, we have }\)
\(\Delta R P Q \sim \Delta R T S\)

Given \(\text { In } \triangle P Q R \angle 1=\angle 2\)
\(\mathrm{PR}=\mathrm{QP}\)
[\(\because\) In a triangle, sides opposite to equal angles are equal]
\(\text { Given } \frac{Q R}{Q S}=\frac{Q T}{P R}\)
\(\text { From (1) and (2) }\)
\(\frac{Q R}{Q S}=\frac{Q T}{Q P} \)
\(\frac{Q S}{Q R}=\frac{Q P}{Q T} \)
\(\text { Now in } \triangle P Q S \text { and } \Delta T Q R\)
\(\frac{Q S}{Q R}=\frac{Q P}{Q T}\)
\(\angle S Q P=\angle R Q T=\angle 1[\text { From (3) }]\)
\(\therefore \text { Using SAS similarity criteria }\)
\(\triangle P Q S \sim \triangle T Q R\)

\(\text { We have } \angle B O C=125^{\circ} \text { and }\)
\(\angle C D O=70^{\circ}\)
\(\text { Since } \left.\angle D O C+\angle B O C=180^{\circ} \text { [linear pair }\right]\)
\(\angle D O C=180^{\circ}-125^{\circ}=55^{\circ}\)
\(\text { In } \triangle D O C \text { Using angle sum property, we get }\)
\(\angle D O C+\angle O D C+\angle D C O =180^{\circ} \)
\(55^{\circ}+70^{\circ}+\angle D C O =180^{\circ} \)
\(\angle D C O =180^{\circ}-55^{\circ}-70^{\circ} \)
\(\angle D C O=55^{\circ}\)
\(\text { Again } \ \triangle O D C \sim \triangle O B A \ \text { (given) }\)
\(\therefore \text { Their corresponding angles are equal. }\)
\(\angle O C D=\angle O A B = 55^{o}\)
\(\therefore \text { From }(1),(2) \text { and (3) }\)
\(\angle D O C=55^{\circ} \)
\(\angle D C O=55^{\circ} \text { and } \)
\(\angle O A B=55^{\circ} \)

We have DB = 3 CD.
BC = BD + DC
BC = 3CD + CD
BC = 4CD
\(CD=\cfrac { 1 }{ 4 } BC\)
\(CD=\cfrac { 1 }{ 4 } BC\)
\(BD=3cD=\cfrac { 3 }{ 4 } BC\)
Since  \(\Delta ABD\) is a right triangle (i) right angled at D.
AB² = AD² + BD²
By \(\Delta ACD\) is a right triangle right angled at D 
AC² = AD² + CD²
Subtracting equation (iii) from equation (ii), we got
AB² - AC² = BD² - CD²
\(\Rightarrow { AB }^{ 2 }-{ AC }^{ 2 }=\left( \cfrac { 3 }{ 4 } BC \right) ^{ 2 }-\left( \cfrac { 1 }{ 4 } BC \right) ^{ 2 }\)
\((from \ CD=\cfrac { 1 }{ 4 } BC,BD=\cfrac { 3 }{ 4 } BC)\)
(i) \(\Rightarrow { AB }^{ 2 }-{ AC }^{ 2 }=\cfrac { 9 }{ 16 } { BC }^{ 2 }-\cfrac { 1 }{ 16 } { BC }\)
\(\Rightarrow { AB }^{ 2 }-{ AC }^{ 2 }=\cfrac { 1 }{ 2 } { BC }^{ 2 }\)
\(\Rightarrow { AB }^{ 2 }-{ AC }^{ 2 }=\cfrac { 1 }{ 2 } BC{ 2 }^{ 2 }\)
\(\Rightarrow 2({ AB }^{ 2 }-{ AC }^{ 2 })={ BC }^{ 2 }\)
\(\Rightarrow { 2AB }^{ 2 }=2{ AC }^{ 2 }+{ BC }^{ 2 }\)

\(\Delta ACD\sim \Delta ABC\)
So,
\(\cfrac { AC }{ AB } =\cfrac { AD }{ AC } \)
AC²  = AB ·AD
Similarly \(\Delta BCD\sim \Delta BAC\)
So,
\(\cfrac { BC }{ BA } =\cfrac { BD }{ BC } \)
BC²  = BA·BD
From (1) and (2)
\(\cfrac { { BC }^{ 2 } }{ AC^{ 2 } } =\cfrac { BA.BD }{ AB.AD } =\cfrac { BD }{ AD } \)

Through O, draw PQIIBC so that P lies on AB and Q lies on DC
Now, PQ II BC
\(PQ\bot AB\quad PQ\bot OC\)
\(\left( \because \angle B={ 90 }^{ 0 }and\angle C={ 90 }^{ 0 } \right) \)
So, \(\angle BPQ={ 90 }^{ 0 }\quad \angle CQP={ 90 }^{ 0 }\)
Therefore BPQC and APQD are both rectangles. Now from \(\Delta OPB\)
OB² = BP² + OP²
Similarly from \(\Delta OQD\)
OD² = OQ² + DQ²
From \(\Delta OQC\)
OC²  = OQ² + CQ²
\(\Delta OAP\) we have
OA² = AP² +OP²
Adding (1) and (2)
OB² + OD² = BP² + OP² + OQ² + DQ²
(As BP = CQ and DQ = AP)
= CQ² + OP² + OQ² + AP²
= CQ² + OQ² + OP² + AP²
= OC²+ OA
[From (3) and (4)]

Let AB be the ladder and CA be the wall with . the window at A.
Also, BC = 2.5 m and CA = 6 m
From Pythagoras theorem
AB² =  BC²+ CA²
= (2.5)² + (6)²
= 42.25
AB = 6.5
Thus, length at the ladder is 6.5 m.

We are given a right triangle ABC right angled at B.
We need to prove that AC² = AB² + BC²
Let us draw \(BD\bot AC\)
Now,\(\Delta ADB\sim \Delta ABC\)
\(\cfrac { AD }{ DB } =\cfrac { BC }{ AC } \)
(sides are proportional)
Also,
\(\Delta BDC\sim \Delta ABC\)
\(\cfrac { CD }{ BC } =\cfrac { BC }{ AC } \)
CD·AC = BC²     ..(2)
Adding (1) and (2)
AD .AC + CD . AC = AB²+ BC²
AC(AD + CD) = AB² + BC²
AC.AC = AB² + BC²
AC = AB² + BC²

BL and CM are medians at the \(\triangle\)ABC in which
\(A=\angle { 90 }^{ 0 }\)
From \(\triangle\)ABC
BC² = AB² + AC²
(Pythagoras theorem)

From \(\Delta ABL\)
BL² = AL² + AB²
\({ BL }^{ 2 }=\left( \cfrac { { AC }^{ 2 } }{ 2 } \right) +{ AB }^{ 2 }\)
(L is the mid-point at AC)
\({ BL }^{ 2 }=\cfrac { { AC }^{ 2 } }{ 4 } +{ AB }^{ 2 }\)
4BL² = AC² + 4AB²
From \(\Delta CMA\)
CM² = AC² + AM²
\({ CM }^{ 2 }={ Ac }^{ 2 }+\left( \cfrac { AB }{ 2 } \right) ^{ 2 }\)
(M is the mid-point at AB)
\({ CM }^{ 2 }={ AC }^{ 2 }+\cfrac { { AB }^{ 2 } }{ 4 } \)
4CM² = 4AC²+ AB²
Adding (2) and (3), we have
4(BL² + CM²) = 5(AC² + AB²)
4(BL² + CM²) = 5BC²

From \(\Delta ADC\) we have
AC² = AD² + CD²       ....(1)
(Pythagoras theorem)
From \(\Delta ADB\) we have
AB² = AD² + BD²     ...(2)
(Pythagoras theorem)
Subtracting (1) from (2) we have,
AB² - AC² = BD² - CD²
AB² + CD² = BD² + AC²