Question

c) In Fig. 4, O is a point on side MN of ΔLMN such that LO = LN. Show that LM > LO.​

Answer 1

Step-by-step explanation:

Let AB intersect LM at O. We have to prove that AO = BO. Now, ∠i = ∠r …

(1) [∵Angle of incidence = Angle of reflection] ∠B = ∠i [Corresponding angles] …(2)

And ∠A = ∠r [Alternate interior angles] …(3)

From (1), (2) and (3), we get ∠B = ∠A ⇒

∠BCO = ∠ACO In ΔBOC and ΔAOC we have ∠1 = ∠2 [Each = 90o]

OC = OC [Common side] And

∠BCO = ∠ACO [Proved above]

ΔBOC ≅ ΔAOC [ASA congruence rule]

Hence, AO = BO [CPCT]