Question

How to prove that ACD is an equilateral triangle?

DB is the diameter in a circle with the center S, . The line AC goes 90 degrees from the center point M of the line SB. "

Answer 1

Hence `Delta ADC` is an equilateral triangle with sides `sqrt3 r`

Explanation:

Let r be the radius of the circle.

`SM = MB = SB / 2 = r / 2`

`SD = SA = SB / 2 = r`

Triangles SAM & BAM are congruent as

SM = BM = r / 2, AM common & /(SMA) = /(BMA) = 90^0#

`:. SA = BA = r`

#:. Delta SAB is equilateral.

`AM = r sin 60 = (sqrt3/2)r`

`AC = 2 * AM = sqrt3 r`

`AD = sqrt((BD^2 - AB^2) = sqrt ((2r)^2 - r^2) = sqrt (3r^2) = sqrt3 r`

In triangle ADC, DM bisects AC at right angle.

`:. Delta ADC` is isosceles and `AD = AC = sqrt3 r`

Since sides AD, DC & AC are equal and `=sqrt3 r, Delta ADC` is an equilateral triangle.

Answer 2

see explanation

Explanation:

The given conditions are shown in the figure.
Let `r` be the radius of the circle.
`SA=SB=SC=SD=r`,
`SM=1/2*SB=r/2`
`sinx=(SM)/(SA)=(r/2)/r=1/2, => x=sin^-1(1/2)=30^@`
`=> angleASM=angleCSM=90-x=90-30=60^@`
`=> CSA=2*60=120^@`
`=> angleCSD=180-angleCSM=180-60=120^@`,
similarly, `angleASD=180-angleASM=180-60=120^@`
As `SA=SC=SD=r, and angleCSA=angleCSD=angleASD=120^@`,
`DeltaSAC, DeltaSCD, and DeltaSDA` are all congruent,
`=> AC=CD=DA`,
Hence, `DeltaACD` is an equilateral triangle.

Answer 3

DB is the diameter in a circle with the center S, . The line AC goes 90 degrees from the center point M of the line SB. "

Given that the line AC goes 90 degrees from the center point M of the line SB.

So in `DeltaASB,SM=MB andAM_|_SB`,

Hence `DeltaASM~=DeltaABM`

So `AS=ABand/_ASB=/_ABS`

Again `SD=SA` (radius of same circle)
So `/_SDA=/_SAD`

Now being diameter the semicircular `/_BAD=90^@`

So in `DeltaBAD,`

`/_ABD+/_ADB=90^@`

`=>/_ASB+/_ADS=90^@`

`=>/_ADS+/_SAD+/_ADS=90^@`

`=>/_ADS+/_ADS+/_ADS=90^@`

`=>3/_ADS=90^@`

`=>/_ADSor/_ADM=90^@/3=30^@`

Now diameter `DB` intersects chord `AC` at M perpendicularly.
So DB bisects AC at M.
This means `DeltaADM~=DeltaCDM`

Hence `/_CDM=/_ADM=30^@=>/_ADC=60^@`

So `/_DAC=/_DCA=60^@`

So `DeltaACD " is equiangular or equilateral"`