Question

Show that a median of a triangle divides it into two triangle of equal areas
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Answer 1

Answer:

Let ABC be a triangle and Let AD be one of its medians.

In \triangle ABD△ ABD and \triangle ADC△ADC the vertex is common and these bases BD and DC are equal.

Draw AE \perp BC.AE⊥BC.

Now area (\triangle ABD) = \dfrac{1}{2} \times base \times altitude\ of \triangle ADBarea(△ABD)=

2

1

×base×altitude of△ADB

= \dfrac{1}{2} \times BD \times AE=

2

1

×BD×AE

= \dfrac{1}{2} \times DC \times AE (\because BD = DC)=

2

1

×DC×AE (∵BD=DC)

but DC and AE is the base and altitude of \triangle ACD△ACD

= \dfrac{1}{2} \times=

2

1

× base DC \times× altitude of \triangle ACD△ACD

= area \triangle ACD=area△ACD

\Rightarrow area (\triangle ABD) = area (\triangle ACD)⇒area(△ABD)=area(△ACD)

Hence the median of a triangle divides it into two triangles of equal areas.

Answer 2

Answer:

let abc be a triangle with AD as a median

construction. draw AL perpendicular to bc

BD =DC

now BD =DC

1/2BD ×AL = 1/2 DC ×AL

ar(ABD) =are(ADC)

hence median divides triangle in two triangle of equal area