Sent a picture of the solution to the problem (s).
Quadrilateral BCDE is inscribed in circle A as shown. `bar(BD)` divides the quadrilateral into two triangles, `DeltaBCD` and `De
ltaBED`. 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`.
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`.
2 answers:
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.`
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