Question

1. given: da⊥ab, cb⊥ab, da≅cb, m is the mid - point of ab. prove: △adm≅△bcm

Explanation:

Step1: Recall mid - point property

Since $M$ is the mid - point of $\overline{AB}$, we have $\overline{AM}\cong\overline{BM}$.

Step2: Consider right - angle property

Given $\overline{DA}\perp\overline{AB}$ and $\overline{CB}\perp\overline{AB}$, then $\angle DAM = \angle CBM=90^{\circ}$.

Step3: Use given congruence

We are given that $\overline{DA}\cong\overline{CB}$.

Step4: Apply SAS congruence criterion

In $\triangle ADM$ and $\triangle BCM$, we have $\overline{DA}\cong\overline{CB}$, $\angle DAM=\angle CBM$, and $\overline{AM}\cong\overline{BM}$. By the Side - Angle - Side (SAS) congruence criterion, $\triangle ADM\cong\triangle BCM$.

Answer:

$\triangle ADM\cong\triangle BCM$ (by SAS)