If l, m, n are three parallel lines and the transversals $t_{1}$ and $t_{2}$ cut the lines l, m, n at the points A, B, C and P, Q, R as shown in the figure, then:

\frac{A B}{B C} = \frac{P Q}{Q R}

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\frac{A B}{Q R} = \frac{B C}{P Q}

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\frac{A P}{B Q} = \frac{B Q}{C R}

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\frac{A B}{P Q} = \frac{A P}{B Q}

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Solution

The correct option is A $\frac{A B}{B C} = \frac{P Q}{Q R}$
Drop perpendiculars $A D E$ and $P S T$

From the similar triangles $A B D$ and $A C E$

$\frac{A B}{A C} = \frac{A D}{A E} . . . . . . . (i)$

Also from the similar triangles $P S Q$ and $P T B$

$\frac{P Q}{P R} = \frac{P S}{P T} . . . . . . . (i i)$

Since $A D = P S$ and $A E = P T$

$\frac{A B}{A C} = \frac{P Q}{P R}$

Applying dividendo

$\frac{A B}{A C - A B} = \frac{P Q}{P R - P Q} \frac{A B}{B C} = \frac{P Q}{Q R}$

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State true or false:

In the following diagram, lines

l, m

and

n

are parallel to each other. Two transversals

p

and

q

intersect the parallel lines at points

A, B, C

and

P, Q, R

respectively as shown, then:

\frac{A B}{B C} = \frac{P Q}{Q R}

Q. If

b c + q r = c a + r p = a b + p q = - 1

, Then

∣ ∣ ∣ ∣ \begin{matrix} a p & b q & c r a & b & c p & q & r \end{matrix} ∣ ∣ ∣ ∣ =

In the follow diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines l, m and n at points A,B,C and P,Q,R respectively as shown below. Prove that $\frac{A B}{B C} = \frac{P Q}{Q R}$ .

Q. If

b c + q r = c a + r p = a b + p q = - 1

, show that

∣ ∣ ∣ ∣ \begin{matrix} a p & a & p b q & b & q c r & c & r \end{matrix} ∣ ∣ ∣ ∣ = 0

Q. If bc +qr = ca + rp = ab + pq = -1 then the value of

∣ ∣ ∣ ∣ \begin{matrix} a p & a & p b q & b & q c r & c & r \end{matrix} ∣ ∣ ∣ ∣

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If l, m, n are three parallel lines and the transversals t1 and t2 cut the lines l, m, n at the points A, B, C and P, Q, R as shown in the figure, then:

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