In the following circle, $m (a r c L A N) = m (a r c M B P)$ , prove that: $L M ∥ N P$

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Solution

Construction: Join LP.
$m (a r c L A N) = m (a r c M B P)$ .... (1) (Given)
$∴ ∠ L P N = \frac{1}{2} m (a r c L A N)$ ...... (2) inscribed $∠$ theorem
$∴ ∠ P L M = \frac{1}{2} m (a r c M B P)$ ....... (3) inscribed $∠$ theorem
$∴ ∠ L P N = ∠ P L M$ from (1), (2) & (3)
$∴ L M | | N P$ alternate $∠$ s test

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