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Aleonysh [2.5K]
2 years ago
10

In the figure, AB║CD and ∠EIA is congruent to ∠GJB Complete the following statements to prove that ∠IKL is congruent to ∠DLH.

Mathematics
2 answers:
Finger [1]2 years ago
7 0

Here it is given that AB || CD

< EIA = <GJB

Now

∠EIA ≅ ∠IKC and ∠GJB is ≅ ∠ JLD (Corresponding angles)

∠EIA  ≅ ∠GJB then ∠IKC ≅ ∠ JLD (Substitution Property of Congruency)

∠IKL + ∠IKC 180° and ∠DLH +  ∠JLD =180° (Linear Pair Theorem)

So

m∠IKL + m∠IKC = 180°       ....(1)

But ∠IKC  ≅ ∠JLD

m∠IKC = m∠JLD (SUBTRACTION PROPERTY OF CONGRUENCY)

So we have

m∠IKL + m∠JLD = 180°

∠IKL and ∠JLD are supplementary angles.

But ∠DLH and ∠JLD are supplementary angles.

∠IKL ≅ ∠DLH (CONGRUENT SUPPLEMENTS THEOREM)

AveGali [126]2 years ago
6 0
The answers 
1)  <span>D. (TRANSITIVE PROPERTY OF CONGRUENCY)
2)  </span><span>C. (LINEAR PAIR THEOREM)
3)  </span><span>D. TRANSITIVE PROPERTY OF CONGRUENCY) to equations (1) and (2), we get m∠IKL + m∠JLD = 180°.
4)  </span><span>C. (CONGRUENT SUPPLEMENTS THEOREM)</span>
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If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?
kondaur [170]

Equation of the circle is (x+6)^{2}+(y+10)^{2}=20.

Solution:

The endpoints of the diameter of a circle are (–8, –6) and (–4, –14).

Center of the circle = Mid point of the diameter

Mid point formula:

$P(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)

Here, x_1=-8, y_1=-6, x_2=-4, y_2=-14

$P(x, y) =\left(\frac{-8-4}{2}, \frac{-6-14}{2}\right)

$P(x, y) =\left(\frac{-12}{2}, \frac{-20}{2}\right)

$P(x, y) =(-6, -10)

Center of the circle = (–6, –10)

Radius is the distance between center and any endpoint of the diameter.

To calculate the radius using distance formula.

r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

Here, x_1=-6, y_1=-10, x_2=-8, y_2=-6

r=\sqrt{\left(-8-(-6)\right)^{2}+\left(-6-(-10)}\right)^{2}}

r=\sqrt{(-8+6)^{2}+(-6+10)^{2}}

r=\sqrt{(-2)^{2}+(4)^{2}}

r=\sqrt{20} units

The standard form of the equation of a circle is

(x-a)^{2}+(y-b)^{2}=r^{2}, where (a, b) are center and r is the radius.

Here, center = (–6, –10) and r=\sqrt{20}

(x-(-6))^{2}+(y-(-10))^{2}={(\sqrt{20})} ^{2}

(x+6)^{2}+(y+10)^{2}=20

Equation of the circle is (x+6)^{2}+(y+10)^{2}=20.

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Answer:

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Step-by-step explanation:

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