Question

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) (ii) ar (AEDF) = ar (ABCDE)
ODG
:​

Answer 1

Mark it as brainliest......

Answer 2

PROOF:-

(i) Since, ∆ACF and ∆ACB are on the same base AC and between same parallels AC and BF.

$\therefore$$\large\tt\red{A(∆ACF)=A(∆ACB)}$

(ii) Adding both sides by AEDC.

$\therefore$$\small\sf{A(∆ACF)+A(AEDC)=A(∆ACB)+A(AEDC)}$

$\large\tt\red{A(AEDF)=A(ABCDE)}$