Part (i): Find EC

Given:

\[ AD = 1.5 \, cm,\quad DB = 3 \, cm,\quad AE = 1 \, cm \]

Since DE ∥ BC, apply BPT:

\[ \dfrac{AD}{DB} = \dfrac{AE}{EC} \]

Substitute values:

\[ \dfrac{1.5}{3} = \dfrac{1}{EC} \]

Simplify LHS:

\[ \dfrac{1.5}{3} = 0.5 \]

So,

\[ 0.5 = \dfrac{1}{EC} \]

Take reciprocal on both sides:

\[ EC = \dfrac{1}{0.5} \]

\[ EC = 2 \, cm \]

Part (ii): Find AD

Given:

\[ AE = 1.8,\quad EC = 5.4,\quad DB = 7.2 \]

Apply BPT:

\[ \dfrac{AE}{EC} = \dfrac{AD}{DB} \]

Substitute values:

\[ \dfrac{1.8}{5.4} = \dfrac{AD}{7.2} \]

Simplify fraction:

\[ \dfrac{1.8}{5.4} = \dfrac{1}{3} \]

So,

\[ \dfrac{1}{3} = \dfrac{AD}{7.2} \]

Cross multiply:

\[ AD = \dfrac{7.2}{3} \]

\[ AD = 2.4 \]

Parallel lines creating proportional segments

In Fig. 6.18, if LM ∥ CB and LN ∥ CD, prove that \[ \dfrac{AM}{AB} = \dfrac{AN}{AD} \]

Given:

\[ LM \parallel CB \] \[ LN \parallel CD \]

To Prove:

\[ \dfrac{AM}{AB} = \dfrac{AN}{AD} \]

Proof:

In triangle \[ \triangle ABC \]

\[ LM \parallel BC \] (Given)

By BPT,

\[ \dfrac{AM}{MB} = \dfrac{AL}{LC} \]    (1)

In triangle \[ \triangle ADC \]

\[ LN \parallel CD \] (Given)

By BPT,

\[ \dfrac{AN}{ND} = \dfrac{AL}{LC} \]    (2)

From (1) and (2),

\[ \dfrac{AM}{MB} = \dfrac{AN}{ND} \]

Taking reciprocal on both sides:

\[ \dfrac{MB}{AM} = \dfrac{ND}{AN} \]

Add 1 to both sides:

\[ \dfrac{MB}{AM} + 1 = \dfrac{ND}{AN} + 1 \]

Combine terms:

\[ \dfrac{MB + AM}{AM} = \dfrac{ND + AN}{AN} \]

Use segment addition:

\[ MB + AM = AB \] \[ ND + AN = AD \]

Therefore,

\[ \dfrac{AB}{AM} = \dfrac{AD}{AN} \]

Taking reciprocal:

\[ \dfrac{AM}{AB} = \dfrac{AN}{AD} \]

Hence proved.

Parallel lines creating proportional divisions

In Fig. 6.19, DE ∥ AC and DF ∥ AE. Prove that \[ \dfrac{BF}{FE} = \dfrac{BE}{EC} \]

Given:

\[ DE \parallel AC \] \[ DF \parallel AE \]

To Prove:

\[ \dfrac{BF}{FE} = \dfrac{BE}{EC} \]

Proof:

In triangle \[ \triangle ABC \]

\[ DE \parallel AC \] (Given)

By BPT,

\[ \dfrac{BD}{DA} = \dfrac{BE}{EC} \]    (1)

Now consider triangle \[ \triangle BAE \]

\[ DF \parallel AE \] (Given)

By BPT,

\[ \dfrac{BD}{DA} = \dfrac{BF}{FE} \]    (2)

From (1) and (2),

\[ \dfrac{BE}{EC} = \dfrac{BF}{FE} \]

\[ \therefore \dfrac{BF}{FE} = \dfrac{BE}{EC} \]

Hence proved.

Using BPT twice to establish parallelism

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

Given:

\[ DE \parallel OQ \] \[ DF \parallel OR \]

To Prove:

\[ EF \parallel QR \]

Proof:

In triangle \[ \triangle POQ \]

\[ DE \parallel OQ \] (Given)

By BPT,

\[ \dfrac{PD}{DO} = \dfrac{PE}{EQ} \]    (1)

In triangle \[ \triangle POR \]

\[ DF \parallel OR \] (Given)

By BPT,

\[ \dfrac{PD}{DO} = \dfrac{PF}{FR} \]    (2)

From (1) and (2),

\[ \dfrac{PE}{EQ} = \dfrac{PF}{FR} \]

Add 1 to both sides:

\[ \dfrac{PE}{EQ} + 1 = \dfrac{PF}{FR} + 1 \]

Combine terms:

\[ \dfrac{PE + EQ}{EQ} = \dfrac{PF + FR}{FR} \]

Use segment addition:

\[ PE + EQ = PQ \] \[ PF + FR = PR \]

Therefore,

\[ \dfrac{PQ}{EQ} = \dfrac{PR}{FR} \]

Taking reciprocal:

\[ \dfrac{EQ}{PQ} = \dfrac{FR}{PR} \]

Thus, in triangle \[ \triangle PQR \]

points E and F divide sides PQ and PR in the same ratio.

By Converse of BPT,

\[ EF \parallel QR \]

Hence proved.

Using two parallels to prove a third parallel

In Fig. 6.21, A, B and C lie on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR. Show that BC ∥ QR.

Given:

\[ AB \parallel PQ \], \[ AC \parallel PR \]

To Prove:

\[ BC \parallel QR \]

Proof:

In triangle \( \triangle OPQ \),

\[ AB \parallel PQ \quad\text{(Given)}\]

By BPT,

\[ \dfrac{OA}{OP} = \dfrac{OB}{OQ}\tag{1} \]

In triangle \( \triangle OPR \),

\[ AC \parallel PR \] (Given)

By BPT,

\[ \dfrac{OA}{OP} = \dfrac{OC}{OR}\tag{2} \]

From (1) and (2),

\[ \dfrac{OB}{OQ} = \dfrac{OC}{OR} \]

This shows that points B and C divide OQ and OR in the same ratio.

Therefore, in triangle \( \triangle OQR \),

By Converse of BPT,

\[ BC \parallel QR \]

Hence proved.

Midpoint theorem using BPT

Using Theorem 6.1, prove that a line drawn through the midpoint of one side of a triangle, parallel to another side, bisects the third side.

Given:

E is the midpoint of AB ⇒ \( AE = EB \)

EF ∥ BC

To Prove:

F is the midpoint of AC i.e., \( AF = FC \)

Proof:

Since E is midpoint of AB,

\[ AE = EB \]

Divide both sides:

\[ \dfrac{AE}{EB} = 1 \tag{1}\]

In triangle \( \triangle ABC \),

\[ EF \parallel BC \quad\text{(Given)}\]

By Theorem 6.1 (BPT),

\[ \dfrac{AE}{EB} = \dfrac{AF}{FC} \]

Substitute from (1):

\[ \dfrac{AF}{FC} = 1 \]

Therefore,

\[ AF = FC \]

Hence, F is the midpoint of AC.

Hence proved.

Using Theorem 6.2, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side.

Given:

E and F are midpoints of AB and AC respectively.

Therefore,

\( AE = EB \)

\( AF = FC \)

Divide both equalities:

\( \dfrac{AE}{EB} = 1 \)

\( \dfrac{AF}{FC} = 1 \)

Thus,

\( \dfrac{AE}{EB} = \dfrac{AF}{FC} \)

In triangle \( \triangle ABC \),

points E and F divide sides AB and AC in the same ratio.

By Theorem 6.2 (Converse of BPT),

\( EF \parallel BC \)

Hence proved.

ABCD is a trapezium in which AB ∥ DC and diagonals intersect at O. Prove that \( \dfrac{AO}{BO} = \dfrac{CO}{DO} \)

Given:

\[ AB \parallel DC \]

To Prove:

\[ \dfrac{AO}{BO} = \dfrac{CO}{DO} \]

Construction:

Through point O, draw a line EF such that \[ EF \parallel AB \parallel DC \]

Proof:

In triangle \( \triangle ACD \),

\[ EF \parallel DC \]

By BPT,

\[ \dfrac{AO}{OC} = \dfrac{AE}{ED} \tag{1}\]

In triangle \( \triangle ABD \),

\[ EF \parallel AB \]

By BPT,

\[ \dfrac{BO}{OD} = \dfrac{AE}{ED} \tag{2}\]

From (1) and (2),

\[ \dfrac{AO}{OC} = \dfrac{BO}{OD} \]

Rearranging,

\[\dfrac{AO}{BO} = \dfrac{CO}{DO} \]

Hence proved.