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Mkey [24]
3 years ago
11

Point Z is equidistant from the vertices of ΔTUV.

Mathematics
2 answers:
stealth61 [152]3 years ago
8 0

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.

Zepler [3.9K]3 years ago
5 0

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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