Question

ABC is an isosceles Triangle with AB = AC. BD and CE are two medians of the triangle. Prove that BD = CE

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Answer 1

SoLuTiOn :

$\sf \red{ In \: \triangle \: ABC, \: It \: is \: given \: that :}$

$\bullet \sf \: AB = AC$

$\rightarrow \: \sf \: \angle B = \angle C.....(i)$

When AB = AC

$\rightarrow\sf \: \dfrac{1}{2} AB = \dfrac{1}{2} AC$

When D and E are the mid-points of AC and AB.

$\rightarrow \sf \: BE = CD.....(ii)$

$\sf \red{ Thus, \: In \: \triangle \: DCB \: and \: \triangle \: EBC : }$

$\bullet \sf \: BC = CB \: \: (Common \: side)$

$\bullet \sf \: DC = EB \: \: \: \: (From \: eq.ii)$

$\bullet \sf \: \angle \: DCB = \angle \: EBC \: \:( From \: eq. \: i)$

By SAS Congruence Condition :

$\sf \purple{ \triangle \: DCB \cong \triangle \: EBC }$

Corresponding parts of congruent triangles are equal. Thus,

$\rightarrow \sf \: BD = CE$

Hence, Proved!

Answer 2

$\huge \underline{ \rm \red{hello}}.$

Question:-

ABC is an isosceles Triangle with AB = AC. BD and CE are two medians of the triangle. Prove that BD = CE

Solution:-

$\huge \underline{ \rm{given:}}$

AB = AC

Also , BD and CE are two medians

Hence ,

E is the midpoint of AB and

D is the midpoint of CE

Hence ,

1/2 AB = 1/2AC

BE = CD

In Δ BEC and ΔCDB ,

BE = CD [ Given ]

∠EBC = ∠DCB [ Angles opposite to equal sides AB and AC ]

BC = CB [ Common ]

∴ Δ BEC ≅ ΔCDB [ SAS ]

BD = CE [ cpct ]

∴ Proved !!!