Question

line segment ab and cd intersect at m.ac parallel to m is mid point of ab prove that m is mid point of cd.
p please answer fast​

Answer 1

$\huge\mathfrak\red{☟︎︎︎ \: \: ᴀɴsᴡᴇʀ \: \: ✍︎}$

If the line segment AB & CD intersect at M and AC is parallel to DB & M is midpoint of AB, the it is proved that M is a midpoint of CD.

Step-by-step explanation:

It is given that,

AC // DB

Line AB intersect line CD at point M

M is the midpoint of AB i.e., AM = B… (i)

Now, consider ΔAMC and ΔBMD, we have

∠AMC = ∠BMD ... [vertically opposite angles]

∠CAM = ∠DBM .... [alternate angles]

∴ By AA similarity, ΔAMC ~ ΔBMD

We know that the corresponding sides of similar triangles are proportional to each other.

∴ AM/BM = CM/DM BM AM

⇒ 1 = DM / CM

from (i) we have AM = BM

⇒ 1 = DM / CM

⇒ CM = DM

∴ M is a mid point of the line segment CD

Hence proved

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Answer 2

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If the line segment AB & CD intersect at M and AC is parallel to DB & M is midpoint of AB, the it is proved that M is a midpoint of CD.

Step-by-step explanation:

It is given that,

AC // DB

Line AB intersect line CD at point M

M is the midpoint of AB i.e., AM = B… (i)

Now, consider ΔAMC and ΔBMD, we have

∠AMC = ∠BMD ... [vertically opposite angles]

∠CAM = ∠DBM .... [alternate angles]

∴ By AA similarity, ΔAMC ~ ΔBMD

We know that the corresponding sides of similar triangles are proportional to each other.

∴ AM/BM = CM/DM BM AM

⇒ 1 = DM / CM

from (i) we have AM = BM

⇒ 1 = DM / CM

⇒ CM = DM

∴ M is a mid point of the line segment CD

Hence proved