In quadrilateral $ACBD$,$AC = AD$ and $AB$ bisects $\angle A$ (see figure). Show that $\Delta ABC \cong \Delta ABD$. What can you say about $BC$ and $BD$?
(N/A) In quadrilateral $ACBD$,we are given:
$AC = AD$
$AB$ bisects $\angle A$,which implies $\angle CAB = \angle DAB$.
Now,consider $\Delta ABC$ and $\Delta ABD$:
$1$. $AC = AD$ (Given)
$2$. $\angle CAB = \angle DAB$ (Since $AB$ bisects $\angle A$)
$3$. $AB = AB$ (Common side)
By the $SAS$ (Side-Angle-Side) congruence criterion,$\Delta ABC \cong \Delta ABD$.
Since the triangles are congruent,their corresponding parts are equal by $CPCT$ (Corresponding Parts of Congruent Triangles).
Therefore,$BC = BD$.
$AC = AD$
$AB$ bisects $\angle A$,which implies $\angle CAB = \angle DAB$.
Now,consider $\Delta ABC$ and $\Delta ABD$:
$1$. $AC = AD$ (Given)
$2$. $\angle CAB = \angle DAB$ (Since $AB$ bisects $\angle A$)
$3$. $AB = AB$ (Common side)
By the $SAS$ (Side-Angle-Side) congruence criterion,$\Delta ABC \cong \Delta ABD$.
Since the triangles are congruent,their corresponding parts are equal by $CPCT$ (Corresponding Parts of Congruent Triangles).
Therefore,$BC = BD$.
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