### If $D$ is the midpoint of the hypotenuse $AC$ of a right $\triangle ABC$, prove that $BD = \dfrac { 1 } { 2 } AC$.

Step by Step Explanation:
1. Let us plot the right $\triangle ABC$ such that $D$ is the midpoint of $AC$.
2. We need to prove that $BD = \dfrac { 1 } { 2 } AC$.

Let us draw a dotted line from $D$ to $E$ such that $BD = DE$ and a dotted line from $E$ to $C$.

In $\triangle ADB$ and $\triangle CDE$, we have \begin{aligned} & AD = CD && \text{[Given]} \\ & \angle ADB = \angle CDE && \text{[Vertically opposite angles]}\\ & BD = ED && \text{[By construction]} \\ & \therefore \triangle ADB \cong \triangle CDE && \text {[By SAS-criterion]} \end{aligned}
3. As the corresponding parts of congruent triangles are equal, we have \begin{aligned} & AB = CE \text{ and } \angle BAD = \angle ECD \end{aligned}

Also, $\angle BAD \text{ and } \angle ECD$ are alternate interior angles. \begin{aligned} & \therefore CE \parallel AB \end{aligned}
4. Now, $CE \parallel AB$ and $BC$ is a transversal. \begin{aligned} & \therefore \angle ABC + \angle BCE = 180^ \circ && \text{[Co-interior angles]}\\ & \implies 90^ \circ + \angle BCE = 180^ \circ && \text{[As } \triangle ABC \text{ is right-angled triangle]} \\ & \implies \angle BCE = 90^ \circ & \end{aligned}
5. Now, in $\triangle ABC$ and $\triangle ECB$, we have \begin{aligned} & BC = CB && \text{[Common]} \\ & AB = EC && \text{[By step 3]} \\ & \angle CBA = \angle BCE && \text{ [Each equal to } 90^ \circ] \\ & \therefore \triangle ABC \cong \triangle ECB && \text{[By SAS-criterion]} \end{aligned} As the corresponding parts of congruent triangles are equal, we have \begin{aligned} & AC = EB \\ \implies & \dfrac { 1 } { 2 } AC = \dfrac { 1 } { 2 } EB \\ \implies & BD = \dfrac { 1 } { 2 } AC \end{aligned}
6. Thus, $$BD = \dfrac { 1 } { 2 } AC$$ 