# Q : 11     In Fig. $\small 9.27$, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that(i)  $\small ar(ACB)=ar(ACF)$ (ii)  $\small ar(AEDF)=ar(ABCDE)$

M manish

We have a pentagon ABCDE in which BF || AC and CD  is produced to F.

(i) Since $\Delta$ACB and $\Delta$ACF are on the same base AC and between same parallels AC and FB.
$\therefore$  ar($\Delta$ACB) = ar ($\Delta$ACF)..................(i)

(ii) Adding the ar (AEDC) on both sides in  equation (i),  we get

ar($\Delta$ACB) + ar(AEDC) =  ar ($\Delta$ACF) + ar(AEDC)
$\therefore$ $\small ar(AEDF)=ar(ABCDE)$

Hence proved.

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