Q

# In Fig.9.23, E is any point on median AD of a ∆ ABC. Show that ar (ABE) = ar (ACE).

Q : 1     In Fig. $\small 9.23$,  E is any point on median AD of a $\small \Delta ABC$. Show that. $\small ar(ABE)=ar(ACE)$

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We have $\Delta$ABC such that AD is a median. And we know that median divides the triangle into two triangles of equal areas.
Therefore, ar($\Delta$ABD) = ar( $\Delta$ACD)............(i)

Similarly, In triangle $\Delta$BEC,
ar($\Delta$BED) = ar ($\Delta$DEC)................(ii)

On subtracting eq(ii) from eq(i), we get
ar($\Delta$ABD) - ar($\Delta$BED) =
$\small ar(ABE)=ar(ACE)$

Hence proved.

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