Question

polygon wtkgmd is shown, where $m\angle wdm = 90^\circ$, $m\angle wgm = 90^\circ$, $m\angle wgk = 90^\circ$, and $m\angle wtk = 90^\circ$.

complete the statement.
given that $\overline{wm} \cong \overline{kw}$ and $\overline{tw} \cong \overline{dw}$, then it can be concluded that $\boldsymbol{\hspace{5em}}$ because of $\boldsymbol{\hspace{2em}}$.

Explanation:

Step1: Identify right triangles

We have two pairs of right triangles: $\triangle WDM$ & $\triangle WTK$, $\triangle WGM$ & $\triangle WGK$.

Step2: Use given congruent sides

Given $\overline{WM} \cong \overline{KW}$ and $\overline{TW} \cong \overline{DW}$. Also, $\angle WDM = \angle WTK = 90^\circ$, $\angle WGM = \angle WGK = 90^\circ$.

Step3: Prove triangle congruence

For $\triangle WDM$ and $\triangle WTK$: $\overline{DW} \cong \overline{TW}$, $\angle WDM \cong \angle WTK$, and we can infer $\overline{DM} \cong \overline{TK}$ (or use HL: right angle, hypotenuse $\overline{WM} \cong \overline{KW}$, leg $\overline{DW} \cong \overline{TW}$). For $\triangle WGM$ and $\triangle WGK$: $\overline{WM} \cong \overline{KW}$, $\overline{WG} \cong \overline{WG}$, right angles, so HL congruence applies.

Step4: Deduce polygon congruence

From triangle congruences, all corresponding sides and angles of quadrilateral $WDMG$ and $WTKG$ are congruent, so $\text{Quadrilateral } WDMG \cong \text{Quadrilateral } WTKG$.

Answer:

$\text{Quadrilateral } WDMG \cong \text{Quadrilateral } WTKG$; Hypotenuse-Leg (HL) Congruence Theorem (for right triangles)