ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.

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Solution

Given : In $Δ A B C, A B = A C B E a n d C F$ are two medians

To prove : BE = CF

Proof : In $Δ A B E a n d Δ A C F,$

AB = AC (Given)

$∠ A = ∠ A (C o m m o n)$

AE = AF (Half of equal sides)

$∴ Δ A B E ≅ Δ A C F (SAS axiom)$

$∴ B E = C F (c . p . c . t .)$

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ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.

Given : In ΔABC, AB=AC BE and CF are two medians To prove : BE = CF Proof : In ΔABE and ΔACF, AB = AC (Given) ∠A=∠A (Common) AE = AF (Half of equal sides) ∴ Δ ABE≅ ΔACF (SAS axiom) ∴ BE=CF (c.p.c.t.)

Given : In $Δ A B C, A B = A C B E a n d C F$ are two medians

To prove : BE = CF

Proof : In $Δ A B E a n d Δ A C F,$

AB = AC (Given)

$∠ A = ∠ A (C o m m o n)$

AE = AF (Half of equal sides)

$∴ Δ A B E ≅ Δ A C F (SAS axiom)$

$∴ B E = C F (c . p . c . t .)$