Question

given: quadrilateral abcd is a kite. prove: △aed≅△ced it is given that quadrilateral abcd is a kite. we know that ad = cd by the definition of. by the kite diagonal theorem, ac is to bd. this means that angles aed and ced are right angles. we also see that ed = ed by the property. therefore, we have that △aed≅△ced by.

Explanation:

Step1: Recall kite - definition

In a kite, two pairs of adjacent sides are equal. So, by the definition of a kite, $\overline{AD}=\overline{CD}$.

Step2: Apply kite - diagonal theorem

The diagonals of a kite are perpendicular. So, $\overline{AC}\perp\overline{BD}$, which means $\angle AED = \angle CED=90^{\circ}$.

Step3: Identify common side

$\overline{ED}$ is common to both $\triangle AED$ and $\triangle CED$. By the reflexive property, $\overline{ED}=\overline{ED}$.

Step4: Apply congruence criterion

We have a right - angle ($\angle AED=\angle CED$), a common side ($\overline{ED}$) and a pair of equal hypotenuses ($\overline{AD}=\overline{CD}$). So, by the Hypotenuse - Leg (HL) congruence criterion for right - triangles, $\triangle AED\cong\triangle CED$.

Answer:

$\triangle AED\cong\triangle CED$ by the Hypotenuse - Leg (HL) congruence criterion.