If one acute angle is double than the other in a right-angled triangle, prove that the hypotenuse is double the smallest side.
Answer:
AC=2BC
- Let △ABC be the right-angled triangle with ∠B=90∘ and ∠ACB=2∠CAB.
Let ∠CAB be equal to x∘. So, ∠ACB=2×∠CAB=2×x∘=2x∘. - We see that AC is the hypotenuse of △ABC.
Also, the side opposite to the smallest angle is the smallest. Thus, AC is the smallest side.
Now, we need to prove AC=2BC. - Let us extend CB to D such that CB=BD and join point A to point D.
- In △ABC and △ABD, we have ∠ABC=∠ABD=90∘[∠ABD=90∘by linear pair]AB=AB[Common]BC=BD[By construction]∴ △ABC≅△ABD[By SAS-criterion]
- As corresponding parts of congruent triangles are equal, we have AC=AD and ∠DAB=∠CAB=x∘ ∴ ∠DAC=∠DAB+∠CAB=x∘+x∘=2x∘
- Now, in △ACD, we have ∠DAC=∠ACD=2x∘⟹AD=CD[Sides opposite to equal angles are equal.]⟹AC=CD[As AD=AC]⟹AC=2BC[As CD=2BC]
- Hence, AC=2BC.