Consider TUV in the figure below.The perpendicular bisectors of its sides are AD, BD, and CD. They meet at a single point D.(In other words, D is the circumcenter of TUV.)Suppose =CD54, =UV148, and =UD90.Find BV, VD, and TC.Note that the figure is not drawn to scale.

Consider XYZ in the figure below.The perpendicular bisectors of its sides are TS, US, and VS. They meet at a single point S.(In other words, S is the circumcenter of XYZ.)Suppose =VS72, =YZ80, and =ZS120.Find XS, UZ, and VY.

In the given figure A, B, C and D are collinear points. If AC = 3 cm, BD = 4 cm andBC = 2 cm, then AD is

Consider GHJ in the figure below.The perpendicular bisectors of its sides are KN, LN, and MN. They meet at a single point N.(In other words, N is the circumcenter of GHJ.)Suppose =LN48, =GK88, and =GN102.Find HN, LJ, and GJ.

Perpendicular bisector passes through the Blank 1 Question 14 of the given segment.

ABCD is a trapezium in which AB||DC and AD = BC. If P, Q, R, S be respectively the mid points of BA, BD, CD and CA, then PQRS is a