Question

Point Z is equidistant from the vertices of ΔTUV.

Which must be true?

TA=TB
AZ=BZ
BTZ=BUZ
TZA= TZB

Answer 1

(C)

Explanation:

It is given that Point Z is equidistant from the vertices of ΔTUV, therefore ZT=ZU=ZV.

Now, from ΔBTZ and ΔBUZ, we have

ZT=ZU (Given)

BZ=ZB (common)

therefore, by RHS rule of congruency,

ΔBTZ≅ΔBUZ

Thus, by corresponding parts of congruent triangles, we have

∠BTZ=∠BUZ

Thus, option C is correct.

Also, it is not necessary that TA=TB and AZ=BZ because there is no information given regarding these equalities.

Answer 2

Consider right triangles BTZ and BUZ.

ZB = ZB (Common)

ZT = ZU (Given)

Therefore, ΔBTZ ≅ ΔBUZ (By RHS theorem)

Hence, by CPCTC,

∠BTZ = ∠BUZ