Point Z is equidistant from the vertices of ΔTUV.

Which must be true?

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

Question

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Respuesta :

JannetPalos JannetPalos · Answer 1 · 2018-10-07T20:08:14+02:00

Consider right triangles BTZ and BUZ.

ZB = ZB (Common)

ZT = ZU (Given)

Therefore, ΔBTZ ≅ ΔBUZ (By RHS theorem)

Hence, by CPCTC,

∠BTZ = ∠BUZ

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boffeemadrid boffeemadrid · Answer 2 · 2019-04-08T16:17:22+02:00

Answer:

(C)

Step-by-step 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.