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
AB=CE and ∠BAD=∠ECD
Also, ∠BAD and ∠ECD are alternate interior angles. ∴CE∥AB - Now, CE∥AB and BC is a transversal. ∴∠ABC+∠BCE=180∘[Co-interior angles]⟹90∘+∠BCE=180∘[As △ABC is right-angled triangle]⟹∠BCE=90∘
- Now, in △ABC and △ECB, we have BC=CB[Common]AB=EC[By step 3]∠CBA=∠BCE [Each equal to 90∘]∴△ABC≅△ECB[By SAS-criterion] As the corresponding parts of congruent triangles are equal, we have AC=EB⟹12AC=12EB⟹BD=12AC
- Thus, BD=12AC