In a right angled triangle, one acute angle is double the other. Prove that the hypotenuse is double the smallest side. GIVEN : A `A B C` in which `/_B=90^0` and `/_A C B=2/_C A B` . to prove : `A C=2B C` CONSTRUCTION : Produce `C B` to `D` such that `B D=C Bdot` Join `A Ddot`

Explanation

Given: `/_\ABC` in which `/_B=90^@` and `/_ACB=2/_CAB`.
To prove: `AC=2BC`
Construction: Produce `CB` to `D` such that `BD=CB`. Join `AD`
Proof : In `/_\ABD` and `/_\ABC`,
`BD=BC` [By construction]
`AB=AB` [Common]
`/_ABD=/_ABC=90^@`
By `SAS` criterion of congruence, we get:
`/_\ABD~=/_\ABC`
`AD=AC` and `/_DAB=/_CAB` [By `CPCT`]
`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` [since `DC=2BC`]
`2BC=AC` [`AD=AC`]
Hence, Hypotenuse is twice the smallest angle.