Given:
AB ∥ CD , BC ∥ AD , BD ∩ AC = O, M∈BC ; N∈AD
Prove: BM = ND
BD and AC are diagonals
MN goes through the center which is also where BD and AC bisect and point O
Can U Guys Put the Answer it would be a BIG HELP, Thanks (:
it is easy to show that
∆NOD ≅ ∆MOB
check out vertical angles and recall that the diagonals bisect each other.
To prove that BM = ND, we can use the property of similar triangles.
First, let's establish some relationships based on the given information:
Since AB ∥ CD, we have triangle AOD ∼ triangle BOC (by corresponding angles formed by parallel lines). Similarly, triangle BOD ∼ triangle AOC (by corresponding angles).
Now, let's consider triangle MOD and triangle NOC.
Since BD and AC are diagonals, they intersect at point O. Therefore, both triangles MOD and NOC share the same height, which is the perpendicular distance between BD and AC. Since both triangles have the same height, we can focus on their bases to prove BM = ND.
In triangle MOD, the base is BD.
In triangle NOC, the base is AC.
Since triangle AOD ∼ triangle BOC, we can establish the following proportion:
BD/OC = AD/AO
Similarly, since triangle BOD ∼ triangle AOC, we have:
AC/OC = CD/AO
Let's rewrite these proportions as:
BD = (OC * AD) / AO -------(1)
AC = (OC * CD) / AO -------(2)
Now, let's consider the ratio of the lengths of BM and ND.
BM/ND = BM/OC + ND/OC
Since BM = BC - CM and ND = AD - AN, we can substitute these values into the ratio:
BM/ND = (BC - CM)/OC + (AD - AN)/OC
From the given information, we know that BC ∥ AD. Therefore, by alternate interior angles, we have:
∠BCM = ∠BDA
Similarly, since AB ∥ CD, we can establish:
∠ADN = ∠ACD
Now, let's rearrange terms in the ratio:
BM/ND = BC/OC - CM/OC + AD/OC - AN/OC
Substituting the proportions (1) and (2) into the equation, we have:
BM/ND = (BD/AO) - (CM/OC) + (AC/AO) - (AN/OC)
Since OC is a common factor in each term, we can factor it out:
BM/ND = (BD - CM + AC - AN)/(AO * OC)
Now, let's consider △MOD and △NOC. We know that angles OMD and ONC are corresponding angles, so they are equal. Therefore, we can conclude that angles MOC and NOC are congruent.
By the Angle Bisector Theorem, we know that MN/MC = ND/DC and MN/NC = BM/MB. However, since BC ∥ AD, we know that DC = MB. Therefore, we have:
MN/MC = ND/MB and MN/NC = BM/MB
By solving these proportions, we can conclude that ND = MN(BC/MC) and BM = MN(BC/NC).
Since ND and BM both equal MN multiplied by a constant (BC/MC and BC/NC, respectively), we can conclude that BM = ND.
Hence, we have proven that BM = ND using the property of similar triangles in conjunction with the given information.