We need to divide this line segment AB of length 14 cm internally in the ratio 2 : 5.
Step 1: Draw a line segment AC of arbitrary length and at an any angle to AB such that ∠CAB is acute.
Step 2: We plot (2 + 5 =) 7 points A1, A2, A3, A4, A5, A6, and A7 such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7
Step 3: We join points A7 and B.
Step 4: We draw line segment A2P such that A2P || A7B and P is the point of intersection of this line segment with AB.
Point P divides AB in the ratio 2 : 5.
In ΔAA2P and ΔAA7B,
v. ∠A is common.
vi. ∠AA2P = ∠AA7B (corresponding angles ∵ A2P || A7B)
Hence, ΔAA2P ~ ΔAA7B
So, ratio of lengths of corresponding sides must be equal.
Let AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = x
So, the previous relation can be re – written as –
⇒ 2(AP +PB) = 7AP
⇒ 2PB = 5AP
⇒ AP/PB = 2/5, or, AP : PB = 2 : 5
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