Question

In a right angled triangle , one acute angle is double the other . Prove that the
hypotenuse is double the smallest side .

Answer 1

Step-by-step explanation:

1. Let ∆ABC is a right triangle such that ∠B = 900 and ∠ACB = 2∠CAB, then we have to prove AC = 2BC. we produce CB to D such that BD = CB and join AD. Proof : In ∆ABD and ∆ABC we have BD = BC [By construction] AB = AB [Common] ∠ABD = ∠ABC = 900 ∴ By SAS criterion of congruence we get ∆ABD ≅ ∆ABC ⇒ AD = AC and ∠DAB = ∠CAB [By cpctc] ⇒ AD = AC and ∠DAB = x [∴∠CAB = x] Now, ∠DAC = ∠DAB + ∠CAB = x + x = 2x ∴ ∠DAC = ∠ACD ⇒ DC = AD [Side Opposite to equal angles] ⇒ 2BC = AD [∵DC = 2BC] ⇒ 2BC = AC [AD = AC] Read more on Sarthaks.com - https://www.sarthaks.com/75118/right-angled-triangle-acute-angle-double-other-prove-that-hypotenuse-double-smallest-side