In Fig. X and Y are the mid-point of AC and AB respectively, QP||BC a

1.	In Fig. X and Y are the mid-point of AC and AB respectively, QP\|\|BC and CYQ and BXP are straight lines. Prove that : ar(Δ ABP) = ar(Δ ACQ).
Answer» In a ΔAXP and ΔCXB, ∠PAX = XCB (Alternative angles AP \|\| BC) AX = CX (Given) ∠AXP = ∠CXB (Vertically opposite angles) ΔAXP ≅ ΔCXB (By ASA rule) AP = BC (By c.p.c.t) ...(i) Similarly, QA = BC ...(ii) From (i) and (ii), we get AP = QA Now, AP \|\| BC And, AP = QA Area (ΔAPB) = Area (ΔACQ) (Therefore, Triangles having equal bases and between the same parallels QP and BC)

In Fig. X and Y are the mid-point of AC and AB respectively, QP||BC and CYQ and BXP are straight lines. Prove that : ar(Δ ABP) = ar(Δ ACQ).

Answer»

In a ΔAXP and ΔCXB,

∠PAX = XCB

(Alternative angles AP || BC)

AX = CX (Given)

∠AXP = ∠CXB

(Vertically opposite angles)

ΔAXP ≅ ΔCXB (By ASA rule)

AP = BC (By c.p.c.t) ...(i)

Similarly,

QA = BC ...(ii)

From (i) and (ii), we get

AP = QA

Now,

AP || BC

And,

AP = QA

Area (ΔAPB) = Area (ΔACQ)

(Therefore, Triangles having equal bases and between the same parallels QP and BC)