f.3 maths(1 question)

Add line AD
Let G be the intersection point of AD and BE
BG = 10/2 = 5cm (mid-pt thm.)
EG = 16/2 = 8cm (mid-pt thm.)
so BE = BG + EG = 5 + 8 = 13cm

mid-pt thm. = mid-point theorem
Let triangle ACD of the diagram as an example
if AB=BC, BG//CD
then 2BG=CD

Join AD w./ st. line & AD intersects BE at X,

BX = 1/2 CD = 5cm (mid-point theorem);
XE = 1/2 AF = 8cm (mid-point theorem);
BE = BX+XE = 13cm

Contruct line DG which is parellel to AB, and intersect BE at O.
AF//BE//CD (given)
AB = BC (given)
AB / BC = FE / ED (intercept theorem)
1 = FE / ED
FE = ED
AB / BC = GO / OD (intercept theorem)
1 = GO / OD
GO = OD

BC//DO (given)
BE//CD (given)
BCDO is a //gram.
BO = 10 cm (opp. sides of //gram)

AC//DG (given)
AG//CD (given)
ACDO is a //gram.
AG = 10 cm (opp. sides of //gram)

In triangle DGF
DO = OG (proved)
DE = DF (proved)
OE = 1/2 GF (mid-point theorem)
OE = 1/2 (AF - AG)
OE = 1/2 (16 - 10)
OE = 3 cm

BE = BO + OE
BE = (10 + 3) cm
BE = 13 cm