Discussion

Explanation:

Draw a line from P parallel to BL which meets AC at M.

Now, let’s consider ∆APM and ∆AQL

∠QAL is common in both triangles, and

∠AQL = ∠APM (Corresponding angles ∵ QL || PM)

∴ ∆APM ∼ ∆AQL ...(By A-A test of similarity)

⇒ AQ/AP = AL/AM

⇒ 1/2 = AL/AM

⇒ AL = LM   …(1)

Similarly, ∆CPM ~ ∆CBL, and

⇒ CM = LM   …(2)

From (1) and (2), we get

CM = ML = LA = 1/3 × AC

Hence, 1.

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