In the figure given below, straight lines $A B$ and $C D$ intersect at $P$ , and $A C | | B D$ . If $B D = 2.4 c m$ , $A C = 4.8 c m$ ; find the value of $\frac{A P}{P B}$ .

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Solution

Given: $A C ∥ B D$
$B D = 2.4 c m$ and $A C = 4.8 c m$

In $△ A C P$ and $△ B D P$
$∠ A C P = ∠ B D P$ [alternate angles]
$∠ C A P = ∠ D B P$ [alternate angles]
and $∠ A P C = ∠ B P D$ [vertically opposite angles]

So, by $A A A$ similarity
$△ A C P$ and $△ B D P$ are similar which means their sides are proportional.

So, $\frac{A C}{B D} = \frac{A P}{B P} = \frac{C P}{P D} . . . . . (i)$

Now,
$\Rightarrow \frac{A C}{B D} = \frac{A P}{B P}$
$\Rightarrow \frac{4.8}{2.4} = \frac{A P}{P B}$
$\Rightarrow \frac{A P}{P B} = 2$

Hence, the required value is $2$ .

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In the figure given below, straight lines AB and CD intersect at P, and AC||BD. If BD=2.4 cm, AC=4.8 cm; find the value of APPB.

In the figure given below, straight lines $A B$ and $C D$ intersect at $P$ , and $A C | | B D$ . If $B D = 2.4 c m$ , $A C = 4.8 c m$ ; find the value of $\frac{A P}{P B}$ .