Given: δwxy is isosceles with legs wx and wy; δwvz is isosceles with legs wv and wz. prove: δwxy ~ δwvz

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

fa tha first one you gonna add y to both sides

it look like this

x-y+y=6+y

x=y+6

therefore your answer is

x =y+6

fa tha second one

you gonna flip tha equation

it gonna look like x-4y=y

then add 4 to both sides

it gonna look like tha

x-4 +4 =y +4

so therefore tha answer is x=y+4

step-by-step explanation: