Let triangle ABC inscribed in (O), point D on arc AB. Through D draw the cord DD'//BC intersect AC at S. The line AD' intersects BC at E. Prove that triangle ADS is similar to triangle ABD '

Clearly, ABCQ and ARBC are parallelograms.

∴BC=AQ and BC=AR

⇒AQ=AR

⇒A is the mid-point of QR.

Similarly, B and C are the mid-points of PR and PQ respectively.

∴AB=

2

1

PQ,BC=

2

1

QR and CA=

2

1

PR

⇒PQ=2AB,QR=2BC and PR=2CA

⇒PQ+QR+RP=2(AB+BC+CA)

⇒ Perimeter of ΔPQR=2 (Perimeter of ΔABC)

there are no figures how am i supposed to answer this question

step-by-step explanation:

hi there!

the answer to this question is d: 35

step-by-step explanation:

5 to the power of 2 is 25

2 multiplied by 5 is 10

then you add the 2 numbers and you get 35