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# Exercise 6.2

In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

(i)

Let EC = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

$\frac{AB}{DB}=\frac{AE}{EC}$

$\frac{1.5}{3}=\frac{1}{x}$

$x=2,&space;EC=2cm$

It is given that DE || BC.

By using basic proportionality theorem, we obtain

$\frac{AD}{DB}=&space;\frac{AE}{EC}$

$\frac{x}{7.2}=\frac{1.8}{5.4}$

$x=2.4&space;\therefore&space;AD=2.4cm$

E and F are points on the sides PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF || QR :

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

(i)

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

$\frac{PE}{EQ}=\frac{3.9}{3}=1.3$

$\frac{PF}{FR}=\frac{3.6}{2.4}=1.5$

$\therefore&space;\frac{PE}{EQ} eq&space;\frac{PF}{FR}$

Therefore, EF is not parallel to QR

(ii)

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm

$\frac{PE}{EQ}=\frac{4}{4.5}=\frac{8}{9}$

$\frac{PF}{FR}=\frac{8}{9}$

therefore, EF is parallel to QR.

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

$\frac{PE}{EQ}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64}$

$\frac{PF}{PR}=\frac{9}{64}$

Hence $\frac{PE}{PQ}=\frac{PF}{PR}$

Therefore, EF is parallel to QR.

In Fig., DE || OQ and DF || OR. Show that EF || QR.

In Δ POQ, DE || OQ

$\therefore&space;\frac{PE}{EQ}=\frac{PD}{DO}.............(i)$ (basic proportionality theorem)

$In&space;\Delta&space;POR,&space;DF&space;||&space;OR$

$\therefore&space;\frac{PF}{FR}=\frac{PD}{DO}$$...............(ii)&space;BPT$

From (i) and (ii) we get,

$\frac{PE}{EQ}=\frac{PF}{FR}$

$\therefore&space;EF\parallel&space;QR$ (converse of BPT)

In Fig., A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

In Δ POQ, AB || PQ

$\therefore&space;\frac{OA}{AP}=\frac{OB}{BQ}.............(i)$

$In&space;\Delta&space;POR,&space;AC&space;||&space;PR$

$\therefore&space;\frac{OA}{AP}=\frac{OC}{CR}..........(ii)$

From (i) and (ii) we get

$\frac{OB}{BQ}=\frac{OC}{CR}$

$\therefore&space;BC\parallel&space;QR$

Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.

i.e., AP = PB and AQ = QC

It can be observed that $\frac{AP}{PB}=\frac{1}{1}$  and $\frac{AQ}{QC}=\frac{1}{1}$

$\therefore&space;\frac{AP}{PB}=\frac{AQ}{QC}$

Hence, by using basic proportionality theorem, we obtain

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $\frac{AO}{BO}=\frac{CO}{DO}$

Draw a line EF through point O, such that  $EF\parallel&space;CD$

In ΔADC, $EO\parallel&space;CD$

By using basic proportionality theorem, we obtain

$\frac{AE}{ED}=\frac{AO}{OC}..................(i)$

In  $\Delta&space;ABD,&space;OE\parallel&space;AB$

So, by using basic proportionality theorem, we obtain

$\frac{ED}{AE}=\frac{OD}{BO}$

$\Rightarrow&space;\frac{AE}{ED}=\frac{BO}{OD}$

$\Rightarrow&space;\frac{AO}{BO}=\frac{OC}{OD}$

The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{AO}{BO}=\frac{CO}{DO}$⋅ Show that ABCD is a trapezium.

Consider the following figure

Draw a line OE || AB

In ΔABD,$OE&space;||&space;AB$

By using basic proportionality theorem, we obtain

$\frac{AE}{ED}=\frac{AO}{OC}$

⇒ EO || DC [By the converse of basic proportionality theorem]

⇒ AB || OE || DC

⇒ AB || CD

∴ ABCD is a trapezium.

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