In fig., ABCDE is a pentagon. A line through B parallel to AC meets DC

1.	In fig., ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that(i) ar.(∆ACB) = ar.(∆ACF) (ii) ar.(∆EDF) = ar.(ABCDE).
Answer» Data: ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. To Prove: (i) ar.(∆ACB) = ar.(∆ACF) (ii) ar.(AEDF) = ar.(ABCDE) Proof: (i) ∆ACB and ∆HCF are on base AC and between AC \|\| BF. ∴ ar.(∆ACB) = ar.(∆ACF) (ii) Now, W(∆ACB) = W(∆ACF) proved, adding ACDE on both sides, ar.(∆ACB) + ar.(ACDE) = ar.(∆ACF) + ar.(ACDE) ∴ ar.(ABCDE) = ar.(AEDF)

In fig., ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that(i) ar.(∆ACB) = ar.(∆ACF) (ii) ar.(∆EDF) = ar.(ABCDE).

Answer»

Data: ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.

To Prove: (i) ar.(∆ACB) = ar.(∆ACF)

(ii) ar.(AEDF) = ar.(ABCDE)

Proof: (i) ∆ACB and ∆HCF are on base AC and between AC || BF.

∴ ar.(∆ACB) = ar.(∆ACF)

(ii) Now, W(∆ACB) = W(∆ACF) proved,

adding ACDE on both sides,

ar.(∆ACB) + ar.(ACDE) = ar.(∆ACF) + ar.(ACDE)

∴ ar.(ABCDE) = ar.(AEDF)