Question #a40d8

2 Answers
Dec 28, 2016

see explanation.

Explanation:

enter image source here

Let #angleEAD=x, angleADC=y#
Given #AE=EF, => angleEFA=x, => angleBFD=x#

Sine Law :

#DeltaACD, (CD)/sinx=(AC)/siny ----- (1)#
#DeltaBFD, (DB)/sinx=(BF)/sin(180-y)=(BF)/siny-----(2)#

Given #CD=DB, (CD)/sinx=(DB)/sinx, => (1)=(2)#
#=> (AC)/siny=(BF)/siny, (sin(180-y)=siny)#

Hence, #AC=BF# (proved)

Dec 28, 2016

drawn

Given ABC a triangle where [AD] is the median and let the segment line [BE] which meets [AD] at F and [AC] at E If we assume that AE=EF, we are to show that AC=BF.

Construction

AD is extended up to G such that # AD=DG# and the point G is joined with C,D and B.

Proof

Now in quadrilateral #ABGC#, D the point of intersection of two diagonals is the mid point of the diagonals as #CD=DB ("given");AD =DG("by construction")# So quadrilateral #ABGC# is a parallelogram.

In #Delta AEF#
#AF=FE->/_FAE=/_AFE=/_BFG ("vertically opposite")#
#=>/_CAG=/_BFG#

Again #AC |\| BG# and AG intercept
So #/_CAG="alternating"/_AGB=/_BGF#

So #/_BFG=/_BGF->BF=BG#
Again #BG=AC("opposite sides of a parallelogram")#

Hence #AC=BF#