Q1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:

$$ DE\parallel BC$$ To Find EC $$\begin{aligned}\dfrac{AD}{DB}&=\dfrac{AE}{EC}\quad\left( \text{By BPT}\right) \\ \dfrac{1.5}{3}&=\dfrac{1}{EC}\\ EC&=\dfrac{3}{1.5}\\ &=2cm\end{aligned}$$

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TO Find AD $$\begin{aligned}\dfrac{AE}{EC}&=\dfrac{AD}{BD}\quad\left( \text{By BPT}\right) \\ \dfrac{1.8}{5.4}&=\dfrac{AD}{7.2}\\ AD&=\dfrac{1.8\times 7.2}{5.4}\\ &=2.4\end{aligned}$$

Q3. In Fig. 6.18, if LM || CB and LN || CD, prove that \(\dfrac{AM}{AB}=\dfrac{AN}{AD}\)

Solution:

Given: $$\begin{aligned}LM\parallel CB\\ LN\parallel CD\end{aligned}$$ To Prove $$\begin{aligned}\dfrac{AM}{AB}=\dfrac{AN}{AD}\end{aligned}$$ Proof: In \(\triangle ABC\) \[LM\parallel BC \quad\text{( Given)}\] therefore by BPT \[\dfrac{AM}{MB}=\dfrac{AL}{LC}\tag{1}\] In triangle \(\triangle ADC\) \[LN\parallel CD\quad\text{( Given)}\] therefore by BPT \[\dfrac{AN}{ND}=\dfrac{AL}{LC}\tag{2}\] From equation-1 and equation-2 \[\dfrac{MB}{AM}=\dfrac{ND}{AN}\] or \[\dfrac{AM}{MB}=\dfrac{AN}{ND}\] Adding 1 to both sides of the equation $$\begin{aligned} \dfrac{MB}{AM}+1&=\dfrac{ND}{AN}+1\\\\ \dfrac{MB+AM}{AM}&=\dfrac{ND+AN}{AN}\\\\ \dfrac{AB}{AM}&=\dfrac{AD}{AN}\\\\ &\text{or}\\\\ \dfrac{AM}{AB}&=\dfrac{AN}{AD}\\ \end{aligned}$$ Thus Proved

Q4. In Fig. 6.19, DE || AC and DF || AE. Prove that
\(\dfrac{BF}{FE}=\dfrac{BE}{EC}\)

Solution:

Given: $$\begin{aligned}DE\parallel AC\\ DF\parallel AE\end{aligned}$$ To Prove: $$\dfrac{BF}{FE}=\dfrac{BE}{EC}$$

Proof:
In triangle \(\triangle ABC\)

$$ DE\parallel AC\quad\text{ (Given)}$$ therefore by BPT $$\dfrac{BD}{DA}=\dfrac{BE}{EC}\tag{1}$$

In triangle \(\triangle ABE\)

$$ \Delta F\parallel AE\quad\text{(Given)}$$ therefore by BPT $$\dfrac{BD}{DA}=\dfrac{BF}{FE}\tag{2}$$ From equation-1 and equation-2 $$\dfrac{BF}{FE}=\dfrac{BE}{EC}$$ Hence Proved.

Q5. In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

Solution:

Given: $$\begin{aligned}DE\parallel OQ\\ DF\parallel OR\end{aligned}$$ To prove: \[EF\parallel QR\] Proof

In Triangle \(\triangle POQ\)

\[DE\parallel OQ\] Therefore, by BPT \[\dfrac{PE}{EQ}=\dfrac{PD}{DO}\tag{1}\]

In triangle \(\triangle POR\)

\[DF\parallel OR\] Therefore by BPT \[\dfrac{PD}{DO}=\dfrac{PF}{FR}\tag{2}\] From equation-1 and equation-2 \[\dfrac{PE}{EQ}=\dfrac{PF}{FR}\] Adding 1 to both sides of the equation $$\begin{aligned}\dfrac{PE}{EQ}+1&=\dfrac{PF}{FR}+1\\\\ \dfrac{PE+EQ}{EQ}&=\dfrac{PF+FR}{FR}\\\\ \dfrac{PQ}{EQ}&=\dfrac{PR}{FR}\end{aligned}$$ By the converse of BPT \[EF\parallel QR\] Hence Proved.

Q6. In Fig. 6.21, A, B, and C are points on OP, OQ, and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:

Given: $$\begin{aligned} AB\parallel PQ\\ AC\parallel PR\end{aligned}$$ To Prove: \[BC\parallel QR\]

Proof:

In triangle \(\triangle POQ\)

\[ AB\parallel PQ\\ \quad\text{(Given)} \] By BPT\ \[\dfrac{PA}{AO}=\dfrac{QB}{BO}\tag{1}\] In triangle \(\triangle POR\)
By BPT \[\dfrac{PA}{AO}=\dfrac{RC}{CO}\tag{2}\] From equation-1 and equation-2 \[\dfrac{QB}{BO}=\dfrac{RC}{CO}\] By Converse of BPT \[BC\parallel QR\] Hence, Proved.

Q7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Solution:

Given:
E is mid-point of AB
To Prove \[\begin{aligned}&EF\parallel BC\\ &\text{F is mid-point of AC}\end{aligned}\] Proof: \[\begin{align}\dfrac{AE}{EB}=1\tag{1}\\ \text{(E is mid-point)} \end{align}\] \[EF\parallel BC\quad\text{( Given)} \] \[\dfrac{AE}{EB}=\dfrac{AF}{FC}\quad\text{(Theorem 6.1)}\] From equation-(1) $$\begin{aligned}\dfrac{AF}{FC}&=1\\ \Rightarrow AF&=FC \end{aligned}$$ Hence, F is a mid-point of side AC

Q8. Using Theorem 6.2, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Solution:

\(E\) and \(F\) are midpoints on \(AB\) and \(AC\) respectively
therefore,

$$\dfrac{AE}{EB}=1$$ (E being midpoint of AB) $$\dfrac{AF}{FC}=1$$ (F being the midpoint of AC) $$\begin{aligned}\Rightarrow \dfrac{AE}{EB}=\dfrac{AF}{FC}\\ \end{aligned}$$ \[EF\parallel BC\quad\text{(Theorem-6.2)}\] Hence, proved.

Q9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(\dfrac{AO}{BO}=Dfrac{CO}{DO}\)

Solution:

Given: $$ AB\parallel DC$$ To Prove $$\dfrac{AO}{BO}=\dfrac{CO}{DO}$$

Construction:
Draw a line \(EF\) passing through \(O\) and parallel to \(AB\)

Proof: $$DC\parallel EF \quad\scriptsize\text{(By Construction)}$$

In triangle \(\triangle ACD\)

By BPT $$\dfrac{AO}{OC}=\dfrac{AE}{ED}\tag{1}$$

In tringle \(\triangle ABD\)

$$\dfrac{BO}{OD}=\dfrac{AE}{ED}\tag{2}$$ From equation-1 and equation-2 $$\begin{aligned}\dfrac{AO}{OC}&=\dfrac{BO}{OD}\\\\ \Rightarrow \dfrac{AO}{BO}&=\dfrac{CO}{DO}\end{aligned}$$ Hence Proved.

10. The diagonals of a quadrilateral ABCD intersect each other at the point O, such that \(\dfrac{AO}{BO}\dfrac{CO}{DO}) Show that ABCD is a trapezium

Solution:

Given $$\dfrac{AO}{BO}=\dfrac{CO}{DO}$$

To Prove:
ABCD is a trapezium

Construction:
Draw a line E F passing through 0 and parallel to AB

Proof: $$AB\parallel EF$$ $$\dfrac{AO}{BO}=\dfrac{CO}{DO}\quad\text{(Given)}\tag{1}$$ By BPT $$\dfrac{BO}{DO}=\dfrac{AE}{ED}$$ From equation-(1) $$\dfrac{AO}{CO}=\dfrac{AE}{ED}$$ By the Converse of BPT $$EF\parallel DC$$ Hence, $$ DC\parallel AB$$ therefore, ABCD is a trapezium.