$ABC$ is an isosceles triangle with $AB = AC$. $BD$ and $CE$ are its two medians. Show that $BD = CE$.
(N/A) Given: $\triangle ABC$ with $AB = AC$.
$D$ is the mid-point of $AC$ and $E$ is the mid-point of $AB$.
To prove: $BD = CE$.
Proof: In $\triangle ABD$ and $\triangle ACE$:
$1$. $AB = AC$ (Given)
$2$. $\angle BAD = \angle CAE$ (Common angle)
$3$. $AD = AE$ (Since $AB = AC$,their halves are equal,i.e.,$\frac{1}{2}AB = \frac{1}{2}AC$)
By $SAS$ congruence criterion,$\triangle ABD \cong \triangle ACE$.
Therefore,$BD = CE$ by $CPCT$ (Corresponding Parts of Congruent Triangles).
$D$ is the mid-point of $AC$ and $E$ is the mid-point of $AB$.
To prove: $BD = CE$.
Proof: In $\triangle ABD$ and $\triangle ACE$:
$1$. $AB = AC$ (Given)
$2$. $\angle BAD = \angle CAE$ (Common angle)
$3$. $AD = AE$ (Since $AB = AC$,their halves are equal,i.e.,$\frac{1}{2}AB = \frac{1}{2}AC$)
By $SAS$ congruence criterion,$\triangle ABD \cong \triangle ACE$.
Therefore,$BD = CE$ by $CPCT$ (Corresponding Parts of Congruent Triangles).
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