If D is the midpoint of the hypotenuse AC of a right △ABC, prove that BD=12AC.
Answer:
- Let us plot the right △ABC such that D is the midpoint of AC.
- We need to prove that BD=12AC.
Let us draw a dotted line from D to E such that BD=DE and a dotted line from E to C.
In △ADB and △CDE, 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