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Likurg_2 [28]
3 years ago
12

Given: ΔWXY is isosceles with legs WX and WY; ΔWVZ is isosceles with legs WV and WZ. Prove: ΔWXY ~ ΔWVZ

Mathematics
2 answers:
Alex777 [14]3 years ago
8 0

Answer:

By AA

ΔWXY ~ΔWVZ

Step-by-step explanation:

Here WXY is an isosceles triangle with legs WX & WY

So WX = WY

Hence ∠X = ∠Y

So ∠2= ∠3.

Now by angle sum property

∠1 + ∠2+∠3 = 180°

∠1+∠2+∠2=180°

2∠2 = 180° - ∠1     .......(1)

In triangle WVZ

WV = WZ

So ∠V = ∠Z

∠4 = ∠5

Once again by angle sum property

∠1 + ∠4 + ∠5=180°

∠1 + ∠4 + ∠4 = 180°

2∠4 = 180° - ∠1       ...(2)

From (1) & (2)

2∠2 = 2∠4

∠2=∠4

Now ∠W is common to both triangles

Hence by AA

ΔWXY ~ΔWVZ

iogann1982 [59]3 years ago
7 0

Answer:

The correct answers are

1. WX = WY; WV = WZ

2. substitution property

3. SAS similarity theorem

Step-by-step explanation:

Just took the Assignment on edge

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