Ex 9.3, 11

Transcript

Ex 9.3, 11 In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar (ACB) = ar (ACF) Given: A pentagon ABCDE where BF ∥ AC To prove: ar (ACB) = ar (ACF) Proof : ΔACB and ΔACF lie on the same base AC and are between the same parallels AC and BF. ∴ ar(ΔACB) = ar(ΔACF) Ex 9.3, 11 In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (ii) ar (AEDF) = ar (ABCDE) In part(i), we proved that ar(ΔACB) = ar(ΔACF) Adding ar(AEDC) both sides ⇒ ar(ΔACB) + ar(AEDC) = ar(ΔACF) + ar(AEDC) ⇒ Area (ABCDE) = Area (AEDF) Hence proved