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#### 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)

#### Solution

#### Similar questions

In the given figure, E is any point on median AD of a ΔABC. Show that

ar (ABE) = ar (ACE)

In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-

(i) ΔMBC ≅ ΔABD

(ii) ar (BYXD) = 2 ar(MBC)

(iii) ar (BYXD) = ar(ABMN)

(iv) ΔFCB ≅ ΔACE

(v) ar(CYXE) = 2 ar(FCB)

(vi) ar (CYXE) = ar(ACFG)

(vii) ar (BCED) = ar(ABMN) + ar(ACFG)

Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.

XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that

ar (ABE) = ar (ACF)

Show that the diagonals of a parallelogram divide it into four triangles of equal area.

In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).