Question

5. The equal sides BA and CA of an isosceles ABAC
are produced beyond the vertex A to the points
E and F so that AE = AF. Join FB and EC and
show that FB = EC.​

Answer 1

In triangle ABF and AEC ,

AB=AC( given)

AE=AF (given)

angle FAB = angle FAC(vertically opposite angles)

==> triangle ABF ≈ AEC

FB = EC

Answer 2

$\sf\large\red{\underline{ correct \: question : }}$

• The equal sides BA and CA of an isosceles ABC are produced beyond the vertex A to the points E and F so that AE = AF. Join FB and EC . Show that FB = EC.

$\\ \bf\large\red{\underline{given : }}$

• △ABC is isoscales.
• AB = AC.
• AE = AF.

$\\ \bf\large\red{\underline{to \: prove : }}$

• FB = EC.

$\\ \bf\large\red{\underline{properties \: used : }}$

• SAS test of congruency :- If two sides and one of the angle is equal , then the triangles are said to be congruent.
• Angles opposite to equal sides are equal.

$\\ \bf\large\red{\underline{solution : }}$

Given that,

➽ AB = AC.

➽ AE = AF.

From the figure , we can see that ...

☛ AE = AB + BE

☛ AF = AC + CF

But , AE = AF and AB = AC

∴ AB + BE = AC + CF

⇒ AB + BE = AB + CF

• since , AB = AC

⇒ BE = CF -------(1)

Now,

In △CFB and △BEC,

• BC = BC ( common side )

• CF = EB ( by equation (1) )

• Angle BCF = Angle EBC ( angle opposite to equal sides are equal )

∴ △CFB ≌△BEC

• By SAS test.

So, FB = EC

• ( common side of congruent triangle )

(proved)