Read the proof.

given: aeec; bddc

prove: △aec ~ △bdc

triangle a e c is shown. line segment b d is drawn near point c to form triangle b d c.

statement reason
1. aeec; bddc 1. given
2. ∠aec is a rt. ∠; ∠bdc is a rt. ∠ 2. definition of perpendicular
3. ∠aec ≅ ∠bdc 3. all right angles are congruent
4. ? 4. reflexive property
5. △aec ~ △bdc 5. aa similarity theorem
what is the missing statement in step 4?

∠ace ≅ ∠bcd
∠eab ≅ ∠dbc
∠eac ≅ ∠eac
∠cbd ≅ ∠dbc

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Ответ:

pettismayersjoy3le8

16.12.2021

Mathematics

$2 {x}^{2} - 6x - 1 = 0 \\ a = 2 \: \: \: b = - 6 \: \: \: c = - 1 \\ \\ x = \frac{6\binom{ + }{ - } \ \sqrt{ {6}^{2} - 4(2)( - 1)} }{2 \times 2} \\ \\ x = \frac{6 \binom{ + }{ - } \sqrt{44} }{4} \\ x = \frac{6 + \sqrt{44} }{4} \\ \\ or x = \frac{6 - \sqrt{44} }{4}$

I hope that is useful for you :)

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