Question

Quadrilateral BCDE is inscribed in circle A as shown. `bar(BD)` divides the quadrilateral into two triangles, `DeltaBCD` and `DeltaBED`. Which statement is true about the triangles?
 A.  The angle bisectors and the perpendicular bisectors for both triangles intersect at the same point.

B. The angle bisectors of `DeltaBCD` intersect at the same point as those of `DeltaBED`.

C. The perpendicular bisectors of `DeltaBCD` intersect at the same point as those of `DeltaBED`.
 
D. The angle bisectors of `DeltaBCD` intersect at the same point as the perpendicular bisectors of `DeltaBED`.

Answer 1

The answer is C i just reviewed my answers, you're welcome!

Answer 2

Answer with explanation:

→It is given  that, Quadrilateral BCDE is inscribed in circle , and segment BD divides the Quadrilateral into two triangles , ΔBED and Δ B CD.

⇒To determine the center of circle,draw perpendicular bisector of any two chords.BC,CD,DE,BD and EB are chords of the circle.

→The Point where , the the perpendicular bisector of sides of Quadrilateral  meet is center of the circle.

→ So, the perpendicular bisector of sides of ΔBED, that is ,BE,B D, and ED meet at the center of the circle.

→Similarly,  the perpendicular bisector of sides of ΔB CD, that is ,BC,C D, and B D meet at the center of the circle.

Option C: The perpendicular bisectors of `ΔBCD` intersect at the same point as those of ΔBED.`