Question

. In the given figure, ABCD is a parallelogram in which diagonals AC and BD intersect at O. A line segment LM is drawn passing through O. Prove that LO = OM.​

Answer 1

$\huge\fcolorbox{black}{lime}{AnsweR:}$

Given ABCD is a parallelogram and AC,

BD is the diagonals intersecting at O.

Hence OA = OC and OB = OD [Since diagonals bisect each other]

Consider, ΔAOL and ΔCOM ∠AOL = ∠COM [Vertically opposite angles]

OA = OC [Given]

∠OAL = ∠OCM [Alternate angles]

∴ ΔAOL ≅ ΔCOM [By ASA Congruence criterion]

Hence OL = OM (CPCT) .

$\large\boxed{\mathfrak{30\:thanks+follow=inbox!}}$

Answer 2

Answer:

Given ABCD is a parallelogram and AC,

BD is the diagonals intersecting at O.

Hence OA = OC and OB = OD [Since diagonals bisect each other]

Consider, ΔAOL and ΔCOM ∠AOL = ∠COM [Vertically opposite angles]

OA = OC [Given]

∠OAL = ∠OCM [Alternate angles]

∴ ΔAOL ≅ ΔCOM [By ASA Congruence criterion]

Hence OL = OM (CPCT) .