Two circles intersect each other at points A and B. Their common tangent touches the circles at points P and Q as shown in the figure.Then, the angles PAQ and PBQ are supplementary.
True
Given - Two circles intersect each other at A and B A common tangent touches the circles at P and Q. PA, PB, QA and QB are joined.
Construction - Join AB,
Proof - PQ is the tangent and AB is the chord
∴∠QPA=∠PBA (alternate segment) ....(i)
similarly we can prove that
∠PQA=∠QBA …(ii)
Adding (i ) and (ii), we get
∠QPA+∠PQA=∠PBA+∠QBA
But ∠QPA+∠PQA=180∘−∠PAQ= …(iii)
(In ΔPAQ)
and ∠PBA+∠QBA=∠PBQ …(iv)
from (iii) and (iv)
∠PBQ=180∘−∠PAQ
⇒∠PBQ+∠PAQ=180∘
⇒∠PAQ+∠PBQ=180∘
Hence ∠PAQ and ∠PBQ are supplementary.