In Fig. 9.24, 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).
Answer
In ΔABC,
AO is the median. (CD is bisected by AB at O)
∴ ar(AOC) = ar(AOD) --- (i)
also,
In ΔBCD,
BO is the median. (CD is bisected by AB at O)
∴ ar(BOC) = ar(BOD) --- (ii)
Adding (i) and (ii) we get,
ar(AOC) + ar(BOC) = ar(AOD) + ar(BOD)
⇒ ar(ABC) = ar(ABD)
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC.
Show that
(i) BDEF is a parallelogram. (ii) ar(DEF) = 1/4 ar(ABC)
(iii) ar (BDEF) = 1/2 ar(ABC)
Answer
Answer
In ΔABC,
AO is the median. (CD is bisected by AB at O)
∴ ar(AOC) = ar(AOD) --- (i)
also,
In ΔBCD,
BO is the median. (CD is bisected by AB at O)
∴ ar(BOC) = ar(BOD) --- (ii)
Adding (i) and (ii) we get,
ar(AOC) + ar(BOC) = ar(AOD) + ar(BOD)
⇒ ar(ABC) = ar(ABD)
In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]
Answer
Let us draw DN ⊥ AC and BM ⊥ AC.
(i) In ΔDON and ΔBOM,
∠ DNO = ∠ BMO (By construction)
∠ DON = ∠ BOM (Vertically opposite angles)
OD = OB (Given)
By AAS congruence rule,
ΔDON ≅ ΔBOM
∴ DN = BM ... (1)
We know that congruent triangles have equal areas.
∴ Area (ΔDON) = Area (ΔBOM) ... (2)
In ΔDNC and ΔBMA,
∠ DNC = ∠ BMA (By construction)
CD = AB (Given)
DN = BM [Using equation (1)]
∴ ΔDNC ≅ ΔBMA (RHS congruence rule)
⇒ Area (ΔDNC) = Area (ΔBMA) ... (3)
On adding equations (2) and (3), we obtain
Area (ΔDON) + Area (ΔDNC) = Area (ΔBOM) + Area (ΔBMA)
Therefore, Area (ΔDOC) = Area (ΔAOB)
(ii) We obtained,
Area (ΔDOC) = Area (ΔAOB)
⇒ Area (ΔDOC) + Area (ΔOCB) = Area (ΔAOB) + Area (ΔOCB)
(Adding Area (ΔOCB) to both sides)
⇒ Area (ΔDCB) = Area (ΔACB)
(iii) We obtained,
Area (ΔDCB) = Area (ΔACB)
If two triangles have the same base and equal areas, then these will lie between the same parallels.
∴ DA || CB ... (4)
In ΔDOA and ΔBOC,
∠ DOA = ∠ BOC (Vertically opposite angles)
OD = OB (Given)
∠ ODA = ∠ OBC (Alternate opposite angles)
By ASA congruence rule,
ΔDOA ≅ ΔBOC
∴ DA = BC ... (5)
In quadrilateral ABCD, one pair of opposite sides is equal and parallel (AD = BC)
Therefore, ABCD is a parallelogram.
Answer
Since ΔBCE and ΔBCD are lying on a common base BC and also have equal areas, ΔBCE and ΔBCD will lie between the same parallel lines.
∴ DE || BC
Answer
Given,
XY || BC, BE || AC and CF || AB
To show,
ar(ΔABE) = ar(ΔAC)
Proof:
EY || BC (XY || BC) --- (i)
also,
BE∥ CY (BE || AC) --- (ii)
From (i) and (ii),
BEYC is a parallelogram. (Both the pairs of opposite sides are parallel.)
Similarly,
BXFC is a parallelogram.
Parallelograms on the same base BC and between the same parallels EF and BC.
⇒ ar(BEYC) = ar(BXFC) (Parallelograms on the same base BC and between the same parallels EF and BC) --- (iii)
Also,
△AEB and parallelogram BEYC are on the same base BE and between the same parallels BE and AC.
⇒ ar(△AEB) = 1/2ar(BEYC) --- (iv)
Similarly,
△ACF and parallelogram BXFC on the same base CF and between the same parallels CF and AB.
⇒ ar(△ ACF) = 1/2ar(BXFC) --- (v)
From (iii), (iv) and (v),
ar(△AEB) = ar(△ACF)