Question

In an isosceles triangle ABC in Fig. AB=AC, show that BF =FC​

Answer 1

Answer:

ABC is an isosceles triangle (given) AB = AC (given) BE and CF are two medians (given) To prove: BE = CF In △CFB and △BEC CE = BF (Since, AC = AB = AC/2 = AB/2 = CE = BF) BC = BC (Common) ∠ECB = ∠FBC (Angle opposite to equal sides are equal) By SAS theorem: △CFB ≅ △BEC So, BE = CF (By c.p.c.t)

Answer 2

side BF = side FC can be shown in the following steps

Step-by-step explanation:

Given:

ΔABC is an isosceles triangle, AB= AC

To show that:

side BF=side FC

Proof:

According to the tangent theorem,

Any tangents drawn from an external point are congruent.

∴ AE=AG......(tangents from external point A) [1]

 BE=BF......(tangents from the external point B) [2]

 FC=GC......(tangents from the external point C)  [3]

It is given that ΔABC is an isosceles triangle

∴ AB = AC

We subtract AE from both the sides and get

AB- AE= AC-AE

AB-AE = AC-AG........(from [1])

∴ BE=GC......[ A-E-B; A-G-C]

But, BE=BF & GC=FC  from [2] and [3]

∴ BF=FC

Hence, proved