Question

In $\Delta ABD$, AB = AD and AC is perpendicular to BD. State the congruence rule by which $\Delta ACB\cong \Delta ACD$.

Answer 1

$\text{In the given triangle,}AB=AD\text{(Given)}AC\text{is the common side}\angle ACB=\angle ACD={90}^{\circ }$

$\therefore \Delta ACB\cong \Delta ACD$ [ RHS criteria]

Hence, by RHS congruence rule, the two triangles are congruent.