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