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 =CD30, =UV142, and =UD78.Find BV, VD, and TC.

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

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

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.

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