Question

Two circles touch each other externally at point P.APC and BPD are straight lines. Show that :(i) triangle PAB and triangle PCD are similar(ii) AB is parallel to CD.​

Answer 1

Answer:

answer is (1).

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Answer 2

Two circles touch each other externally at point P.APC and BPD are straight lines.

Construction: draw a common tangent from point P and mark it MN.

∠ APN = ∠ B          (∵ MN is a tangent)

Similarly, ∠ CPM = ∠ D

But we have,

∠ APN = ∠ CPM

∴ ∠ B = ∠ D

In Δ PAB and Δ PCD

∠ B = ∠ D

∠ APB = ∠ CPD

∴ Δ PAB ~ Δ PCD      

As ∠ B = ∠ D

and are alternate angles, we have

∴ AB ║ CD