Given: ΔABC and ΔABD are two triangles on the same base AB. Line segment CD is bisected by AB at O.
To Prove: ar(ABC) = ar(ABD)
Proof: ∵ Line segment CD is bisected by AB at O
∴ CO = DO
⇒ O is the mid-point of CD.
⇒ AO is a median of ΔACD and BO is a median of ΔBCD
∵ AO is a median of ΔACD
∴ ar(ΔAOC) = ar(Δ AOD) ... (1)
∵ A median of a triangle divides it into
two triangles of equal areas
∵ BO is a median of ΔBCD ar(ΔBOC) = ar(ΔBOD) ...(2)
∵ A median of a triangle divides it into
two triangles of equal areas
Adding (1) and (2), we get ar(ΔAOC) + ar(ΔBOC)
= ar(ΔAOD) + ar(ΔBOD)
⇒ ar(ΔABC) = ar(ΔABD)
ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that
(i) ar(ΔADO) = ar(ΔCDO)
(ii) ar(ΔABP) = ar(ΔCBP)