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

Step by Step Explanation:
  1. Let ABC be the right-angled triangle with B=90 and ACB=2CAB.

    Let CAB be equal to x. So, ACB=2×CAB=2×x=2x.
      A C B 2x°
  2. 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.
  3. Let us extend CB to D such that CB=BD and join point A to point D.
      A C D B 2x°
  4. In ABC and ABD, we have ABC=ABD=90[ABD=90by linear pair]AB=AB[Common]BC=BD[By construction]
  5. As corresponding parts of congruent triangles are equal, we have \begin{aligned} AC = AD \text{ and } \angle DAB = \angle CAB = x^ \circ \end{aligned} \therefore {\space} \angle DAC = \angle DAB + \angle CAB = x^ \circ + x^ \circ = 2x^ \circ
  6. Now, in \triangle ACD , we have \begin{aligned} & \angle DAC = \angle ACD = 2x^ \circ \\ \implies & AD = CD && \text{[Sides opposite to equal angles are equal.]}\\ \implies & AC = CD && [As \space AD = AC] \\ \implies & AC = 2BC && [As \space CD = 2BC] \end{aligned}
  7. \text{Hence, } \bf {AC = 2BC} .

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