5

enter image description here We have that $DS \parallel BA$ and $∠DOS+∠DTS=180°$ and $O$ is the centre of the circle. I should somehow prove that $AB=AC$. In case you'd tell me I didn't show any effort I want to tell you that I tried to make this drawing for 40 minutes. Thank you for any help you could provide me.

amWhy
  • 209,954
trying to be mathematician
  • 227

1 Answers1

3

If $AB=AC$, using properties of an isosceles triangle, $$\angle ACB=\angle ABC$$so this will be the goal of our proof.

Some headstart for you:

We have $$∠DOS+∠DTS=180^\circ$$ so $$∠TDO+∠TSO=180^\circ\tag{1}$$

Since $DS\parallel BA$, $$∠SDB+\underbrace{∠DBA}_{=\angle ABC}=180^\circ\tag{2}$$

We also have $$∠ABC+∠ACB+∠BAC=180^\circ$$

Your turn :)

Karn Watcharasupat
  • 4,323
  • 1
    What I got was that if $∠DTS=α$, then $∠DOS=∠BOA=180°-α$, and since $BO=OA$, $∠OBA=∠OAB=\frac{α}{2}$ and $BCA=90°-\frac{α}{2}$, so all there's left to prove there is that $∠CBO=90°-α$. Though I haven't used that $DS \parallel BA$. – trying to be mathematician May 03 '18 at 11:34
  • 1
    So since $AO=BO$ and $DS \parallel AB$, can we make a presupmtion that it must be an iscosceles trapezoid? If so, then $∠SDO=∠DSO$ and $∠ABD=∠BAS=∠TDS=∠DST$, therefore $∠TDO=∠TSO$ and since we know that $∠TDO+∠TSO=180°$, we get that both of these are equal to 90°. Therefore, the proof that $∠CBO=90°−α$ comes easily out of the triangle $TSB$, am I right? – trying to be mathematician May 03 '18 at 12:02
  • @tryingtobemathematician Yep you are! Good job! – Karn Watcharasupat May 03 '18 at 18:02
  • Good to know :). Thank you very much! – trying to be mathematician May 03 '18 at 19:46