1. ## Geometry - Help

In this figure Prove $"Angle" RQS = 2 \times "Angle" BDC$

"Angle" QRD = "Angle" SRB,
"Angle" RSD = "Angle" ASC and
DRB, DSC, QSA are straight lines.

2. Let $\widehat{QRS}=\widehat{BRS}=x$ and $\widehat{RSD}=\widehat{ASC}=y$
$\widehat{QRS}=180-2x$
$\widehat{RQS}=180-\widehat{QRS}-\widehat{QSR}=$
$=180-2y-(180-2x)=2(x-y)$

$\widehat{BDC}=\widehat{RDS}=180-\widehat{DSR}-\widehat{SRD}=$
$=180-x-y-(180-2x)=x-y$
So, $\widehat{QRS}=2\widehat{BDC}$

3. Originally Posted by name101
In this figure Prove $"Angle" RQS = 2 \times "Angle" BDC$

"Angle" QRD = "Angle" SRB,
"Angle" RSD = "Angle" ASC and
DRB, DSC, QSA are straight lines.
I'm like, totally in awe of red_dog right now.

How can you do the problem if there is no point R in the figure??

-Dan

4. Originally Posted by name101
DRB, DSC, QSA are straight lines.
DRB is a straight line, so I assumed R is the point on DB.

5. Originally Posted by red_dog
DRB is a straight line, so I assumed R is the point on DB.
Ahhhhhh! I hear the grasshopper now. (Bows in respect.)

-Dan

6. sorry i left that minor detail out. Quick to pick this one up.
Very good work.
Thanks again.