If ^@D^@ is the midpoint of the hypotenuse ^@AC^@ of a right ^@ | Math Question and Answer

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

Answer:

Step by Step Explanation:

Let us plot the right △ABC $△ A B C$ such that D $D$ is the midpoint of AC $A C$ .
We need to prove that BD=12AC $B D = \frac{1}{2} A C$ .

Let us draw a dotted line from D $D$ to E $E$ such that BD=DE $B D = D E$ and a dotted line from E $E$ to C $C$ .
In △ADB $△ A D B$ and △CDE $△ C D E$ , we have AD=CD[Given]∠ADB=∠CDE[Vertically opposite angles]BD=ED[By construction]∴△ADB≅△CDE[By SAS-criterion]
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. $\begin{aligned} & \therefore CE \parallel AB \end{aligned}$
Now, $CE \parallel AB$ and $BC$ is a transversal. $\begin{aligned} & \therefore \angle ABC + \angle BCE = 180^ \circ && \text{[Co-interior angles]}\\ & \implies 90^ \circ + \angle BCE = 180^ \circ && \text{[As } \triangle ABC \text{ is right-angled triangle]} \\ & \implies \angle BCE = 90^ \circ & \end{aligned}$
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 $\begin{aligned} & AC = EB \\ \implies & \dfrac { 1 } { 2 } AC = \dfrac { 1 } { 2 } EB \\ \implies & BD = \dfrac { 1 } { 2 } AC \end{aligned}$
Thus, $BD = \dfrac { 1 } { 2 } AC$

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