In the adjoining figure, D, E and F are midpoints of the sides BC, CA and AB respectively of ∆ABC. Prove that BCEF is a trapezium and area of the trap. BCEF = area of ∆ABC.

Question

KnowledgeBoat · Accepted Answer

We know that D and E are the mid-points of BC and CA, respectively. By mid-point theorem, DE || AB and DE = AB = BF (As F is mid-point of AB) From figure, BF || DE and BF = DE. Hence, BDEF is a || gm. Similarly, F and E are the mid-points of AB and CA. By mid-point theorem, EF || BC and EF = BC = DC (As D is mid-point of BC) From figure, EF || DC and EF = DC. Hence, EFDC is a || gm. Since, FE || BC and FB, EC are not parallel. Hence, EFBC is a trapezium. F and D are the mid-points of AB and BC. By mid-point theorem, FD || AC and FD = AC = AE (As E is mid-point of AC) From figure, FD || AE and FD = AE. Hence, AFDE is a || gm. We know that a diagonal divides a || gm in two triangles of equal area. In || gm BDEF, FD is the diagonal, ∴ area of △DEF = area of △BDF .........(i) In || gm EFDC, DE is the diagonal, ∴ area of △DEF = area of △EDC .........(ii) In || gm AFDE, FE is the diagonal, ∴ area of △DEF = area of △AFE .........(iii) From (i), (ii) and (iii) we get, area of △DEF = area of △BDF = area of △EDC = area of △AFE .......(iv) From figure, area of △ABC = area of △DEF + area of △BDF + area of △EDC + area of △AFE = area of △DEF + area of △DEF + area of △DEF + area of △DEF = 4 x area of △DEF ∴ area of △ABC = 4 x area of △DEF ⇒ area of △DEF = area of △ABC ......(v) From figure, area of trapezium BCEF = area of △DEF + area of △BDF + area of △EDC = area of △DEF + area of △DEF + area of △DEF = 3 x area of △DEF ∴ area of trapezium BCEF = 3 x area of △DEF ⇒ area of △DEF = x area of trapezium BCEF .......(vi) From (v) and (vi) we get, ⇒ area of △ABC = area of trapezium BCEF ⇒ area of trapezium BCEF = area of △ABC. Hence, proved that area of trapezium BCEF = area of △ABC.

Mathematics

In the adjoining figure, D, E and F are midpoints of the sides BC, CA and AB respectively of ∆ABC. Prove that BCEF is a trapezium and area of the trap. BCEF = area of ∆ABC.

Theorems on Area

Related Questions

In figure (1) given below, point D divides the side BC of ∆ABC in the ratio m : n. Prove that area of ∆ABD : area of ∆ADC = m : n.

In the figure (2) given below, P is a point on the side BC of ∆ABC such that PC = 2BP, and Q is a point on AP such that QA = 5PQ, find area of ∆AQC : area of ∆ABC.