Question

Triangles on the same base and between the same parallels are equal in area

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Answer 1

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Triangles on the same base and between the same parallels are equal in area

$\huge\purple{\mid{\fbox{\tt{Solution}}\mid}}$

$\blue{Given}$

• Two ∆ABC and DBC on the same base BC and between the same parallel lines BC and AD.

$\green{To\:prove}$

• ar(∆ABC) = ar(∆DBC).

$\red{Construction}$

Through B, draw BE || CA, meeting DA produced in E. And, through C,draw CF || BD, meeting AD produced in F.

$\pink{Proof}$

BE || CA and EA || BC ⇒ BCAE is a ||gm.

CF || BD and DF || BC ⇒ BCFD is a ||gm.

Clearly,the ||gms BCAE and BCFD are on the same base BC and between the same parallel lines BC and EF.

∴ ar(||gm BCAE) = ar(||gm BCFD)⠀⠀⠀⠀⠀⠀⠀...(i)

But, a diagonal of a ||gm divide it itno two triangles of equal areas.

$\therefore \: ar(\triangle \: ABC = \frac{1}{2} ar( || gm \: BCAE) \: \: \: \: \: \: \: \: \:... (ii)$

$and \: ar(\triangle \: DBC) = \frac{1}{2} ar( || gm \: BCFD). \: \: \: \: \: \: ...(iii)$

Thus, from (i),(ii) and (iii),we get

ar(∆ABC) = ar(∆DBC).

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Answer 2

Answer:

answer ar 24500

Step-by-step explanation:

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