Question

two triangles having the same base and equal areas lie between the same parallels. Prove it

Answer 1

$\huge{Hey\:Mate\:Here\:Is\:Your\:Answer}$

$\textbf{Given That ...}$

∆ABC And ∆ADC Both Lie On The Same Base In A Way That

ar( ∆ABC ) = ar( ∆ABD)

$\textbf{To Prove..}$

∆ABC And ∆ADB Lie Between The Same Parallel.. ie:- CD || AB

$\textbf{Construction}$

Altitude CE And DF Of ∆ACB & ∆ADB.. On AB .

$\textbf{Proof -}$

Now According To Question It's Said That ∆ABC And ∆ABD Both lie On The Same Base ... And Both Have

Equal Area.

Now By Construction We Have ...

CE Perpendicular To AB

And DF Perpendicular To AB

Now We Know That Lines Perpendicular To Same Line Are Parallel To Each Other Therefore ...

CE || DF --- EQ (1)

So Now It's Given That

ar ( ABC ) = ar ( ABD )

Now We Know That ..

$\boxed{Area\:Triangle\:= \:1/2 \:×\: Base\: ×\: Height}$

Therefore

ar(ABC) = 1/2 × AB × CE

And

ar(ABD) = 1/2 × AB × DF

Now Since Area Of ABC = Area OF ABD

Therefore We Have ...

$\boxed{CE = DF}$ ---- EQ. (2)

Now In Quadrilateral CDEF

CE = DF

And

CE || DF

So We Know That If In A Quadrilateral If One Pair Of Opposite Sides Are Equal And Parallel Then The

Quadrilateral Is Parallelogram.

Therefore

CDEF Is A Parallelogram ...

Therefore

CD || EF ( opposite Sides Of Parallelogram )

That Is ..

$\boxed{CD || AB}$

$\textbf{Hence Proved That Triangle ABC And Triangle ADB lie Between The same Parallel ....}$