Japan
If ^@D^@ is the midpoint of the hypotenuse ^@AC^@ of a right ^@ \triangle ABC^@, prove that ^@ BD = \dfrac { 1 } { 2 } AC ^@.
Answer:
- Let us plot the right ^@ \triangle ABC^@ such that ^@D^@ is the midpoint of ^@AC^@.
- 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}@^ - 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 @^
