As, \(\triangle ABC\sim \triangle AMP\) [proved in part(i)]
\(\therefore d\frac{AC}{AP}=\frac{BC}{MP}\)
[since ratio of the corresponding sides of similar triangles are equal]
\(\Rightarrow \frac{CA}{PA}=\frac{BC}{MP}\)

Given, AD and CE are altitudes which intersect each other at the point P.
In ΔPDC and ΔBEC,
∠PDC = ∠BEC           [each 90°]
and ∠PCD = ∠BCE   [Common angle]
\(\therefore\) ΔPDC ∼ ΔBEC     [by AA similarity criterion]

Given, AD and CE are altitudes which intersect each other at the point P.
In ΔAEP and ΔADB,
∠AEP = ∠ADB            [Each 90°]
and ∠PAE = ∠BAD    [Common angle]
\(\therefore\) ΔAEP ∼ ΔADB       [by AA similarity criterion]

Given, AD and CE are altitudes which intersect each other at the point P.
In ΔABD and ΔCBE,
∠ADB = ∠CEB      [each 90°]
and ∠ABD = ∠CBE      [Common angle]
\(\therefore\) ΔABD ∼ ΔCBE    [by AA similarity criterion]

In Δ POQ, DE || OQ    [given]
\(\therefore \frac{P E}{E Q}=\frac{P D}{D O}\) ...(i)
[by basic proportionality theorem] 
In ΔPOR,    DF || OR     [given]
\(\therefore \frac{P F}{F R}=\frac{P D}{D O}\) ...(ii)
[by basic proportionality theorem]
From Eqs. (i) and (ii), we get 
\(\frac{P E}{E Q}=\frac{P F}{F R}\)
In ΔPQR, we have \(\frac{P E}{E Q}=\frac{P F}{F R}\)
∴  EF || QR
[by Converse of basic proportionally theorem]
Hence proved.

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
\( \frac{P E}{P Q}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64} \)
\(\frac{P F}{P R}=\frac{0.36}{2.56}=\frac{9}{64} \)
Hence \(\frac{P E}{P Q}=\frac{P F}{P R}\)
Therefore EF is parallel to QR

Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
\(\frac{A D}{D B}=\frac{A E}{E C}\)
\(\frac{1.5}{3}=\frac{1}{x}\)
\(x=\frac{3 \times 1}{1.5}\)
x =  2
∴ EC = 2 cm

Given, \(\angle ABC={ 90 }^{ ° }\) and \(\angle AMP={ 90 }^{ ° }\)
In \(\triangle ABC\) and \(\triangle AMP\)
\(\angle ABC=\angle AMP\)  [each 90°]
and \(\angle BAC=\angle MAP\) [common angle]
\(\therefore \triangle ABC\sim \triangle AMP\) [by AA similarity criterion]

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
\( \frac{P E}{E Q}=\frac{3.9}{3}=1.3 \)
\(\frac{P F}{F R}=\frac{3.6}{2.4}=1.5\)
Hence, \(\frac{P E}{E Q} \neq \frac{P F}{F R}\)
Therefore, EF is not parallel to QR

Draw \(\triangle PQR\), such that S and T are points on sides PR and QR respectively. We join the points S and T.

From the above figure, we have \(\triangle RPQ\) and \(\triangle RTS\) in which
\(\angle RPQ=\angle RTS\) [given]
\(\Rightarrow\angle PRQ=\angle TRS\) [common angle]
\(\therefore \triangle RPQ\sim \triangle RTS\) [by AA similarity criterion]
Hence proved.

In \(\triangle PQR, \angle 1=\angle 2\)  [given]
\(\Rightarrow \) PR = PQ  [since, sides opposite to equal angles of a triangle are not equal]
Given that, \(\frac { QR }{ QS } =\frac { QT }{ PR } \Rightarrow \quad \frac { QR }{ QS } =\frac { QT }{ PQ } \)
[\(\because \)  PQ = PR, proved above]
\(\Rightarrow \frac { QS }{ QR } =\frac { PQ }{ QT } \)  
[on taking of the terms reciprocals]  ... (i)
In \(\triangle PQS\) and \(\triangle TQR\), we have
\(\angle PQS=\angle TQR\)     [common]
and \(\Rightarrow \frac { QS }{ QR } =\frac { QP }{ QT } \)  [from Eq.(i)]
\(\therefore \triangle PQS\sim \triangle TQR\)    [by SAS similarity criterion]
Hence proved.

Consider \(\triangle ABC\), in which D is the mid-point of AB.
Then, \(\frac { AD }{ DB } =1\)    ... (i)
Line l is drawn through D such that \(l\parallel BC\) and it meets AC at E.
By basic proportionality theorem, 
\(\frac { AD }{ DB } =\frac { AE }{ EC } \)

\(\Rightarrow \frac { AE }{ EC } =1\) [from Eq. (i)]
\(\Rightarrow AE=EC\)
So, E is the mid-point of AC.
Hence, a line drawn through the mid-point of one side of a triangle parallel to another side, bisects the third side.
Hence proved.

Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post. 
From the figure, you can see that DE is the shadow of the girl. Let DE be x metres.

Now, BD = 1.2 m \(\times\) 4 = 4.8 m.
Note that in \(\Delta\) ABE and \(\Delta\) CDE,
\(\angle\)B and \(\angle\)D  (Each is of 90° because lamp-post as well as the girl are standing vertical to the ground)
and \(\angle\)E = \(\angle\)E   (Same angle)
So, \(\Delta \mathrm{ABE} \sim \Delta \mathrm{CDE}\) (AA similarity criterion)
Therefore, \(\frac{\mathrm{BE}}{\mathrm{DE}}=\frac{\mathrm{AB}}{\mathrm{CD}}\)
i.e.,   \(\frac{4.8+x}{x}=\frac{3.6}{0.9} \quad\left(90 \mathrm{~cm}=\frac{90}{100} \mathrm{~m}=0.9 \mathrm{~m}\right)\)
i.e., 4.8 + x = 4x
i.e.,  3x = 4.8
i.e.,  x = 1.6
So the shadow of the girl after walking for 4 seconds is 1.6 m long.

Δ ABC ~ Δ PQR    (Given)
So, \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{CA}}{\mathrm{RP}}\)                                  (1)
and \(\angle\) A = \(\angle\) P, \(\angle\) B = \(\angle\) Q and \(\angle\) C = \(\angle\) R           (2)
But AB = 2 AM and PQ = 2 PN 
                              (As CM and RN are medians)
So, from (1), \(\frac{2 \mathrm{AM}}{2 \mathrm{PN}}=\frac{\mathrm{CA}}{\mathrm{RP}}\)
i.e., \(\frac{\mathrm{AM}}{\mathrm{PN}}=\frac{\mathrm{CA}}{\mathrm{RP}}\)                                            (3)
Also, \(\angle\) MAC = \(\angle\) NPR  [From (2)]                         (4)
So, from (3) and (4),
Δ AMC ~ Δ PNR      (SAS similarity)                       (5)

Draw \(\triangle ABC\) such that D is a point on BC and join AD.

In \(\triangle ABC\) and \(\triangle DAC\) , we have
\(\angle BAC=\angle ADC\)  [given]
and \(\angle ACB=\angle DCA\) [common angle]
\(\therefore \quad \triangle ABC\sim \triangle DAC\) [by AA similarity criterion]
\(\Rightarrow \frac{AC}{DC}=\frac{CB}{CA}\)
[since, corresponding sides of two similar triangles are proportional]
or \(\frac{CA}{CD}=\frac{CB}{CA}\Rightarrow\)  CA x CA = CB x CD
\(\Rightarrow { CA }^{ 2 }=CB\times CD\)
Hence proved.

Given ABCD is a quadrilateral. Its diagonal AC and BD intersect each other at O such that
\(\frac { AO }{ BO } =\frac { CO }{ DO } \Rightarrow \frac { AO }{ CO } =\frac { BO }{ DO } \)    ....(i)
To prove  ABCD is a trapezium
Construction Through O, we draw \(OE\parallel BA\). Let OE meets AD at E.

 In \(\triangle DAB, EO\parallel AB\)  [by construction]
\(\Rightarrow \frac { DE }{ EA } =\frac { DO }{ BO } \) 
[by basic proportionality theorem]
or \(\frac { AE }{ ED } =\frac { BO }{ DO } \)
[on taking reciprocal of the terms] ...(ii)
From Eqs. (i) and (ii), we get
\(\frac { AO }{ CO } =\frac { AE }{ ED } \)
\(\Rightarrow OE\parallel DC\)
[by converse of basic proportionality theorem]
Now, \(BA\parallel OE\) and \(OE\parallel CD\) 
\(\therefore AB\parallel DC\)
[since lines parallel to the same line are also parallel to each other]
Hence, quadrilateral ABCD is a trapezium.
Hence proved.

In \(\triangle OPQ\),  \(AB\parallel PQ\) [given]
\(\therefore \quad \frac { OA }{ AP } =\frac { OB }{ BQ } \)  ... (i)
[by basic proportionality theorem]
Also, in \(\triangle OPR\), \(AC\parallel PR\)  [given]
\(\therefore \quad \frac { OA }{ AP } =\frac { OC }{ CR } \)  ... (ii)
From Eqs. (i) and (ii), 
\(\frac { OB }{ BQ } =\frac { OC }{ CR } \Rightarrow \quad BC\parallel QR\)
[by converse of basic proportionality theorem]
Hence proved.