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3 years ago

The figure below shows a quadrilateral ABCD with diagonal BD bisecting angle ADC:

Mathematics

2 answers:

Neko [114]3 years ago

6 0

The answer would be B) AD=DC, the reason why is because line AD is a reflection of line DC

Alinara [238K]3 years ago

6 0

Answer with explanation:

In Quadrilateral A B CD

Diagonal BD bisects angle D.

→→ ∠A DB=∠ C DB

In ΔA DB and ΔC DB

1. AB= CB→→→→[Given]

2. ∠A DB=∠ C DB-------[Given]

3. BD=BD ----[Common side]

⇒ ΔA DB ≅ ΔC DB→→→[S A S]

AD=DC→→→→[C P C T]

Option B:⇒ AD = DC

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Anni [7]

Answer:

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Step-by-step explanation:

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3 years ago

C is the circumcenter of APQR.<br> PC=3x+7, RC=5x-15, & QC=51-X<br> x=[?]

amm1812

Answer:

x = 11

Step-by-step explanation:

C is the circumcenter of $\triangle PQR$

$\implies PC=QC=RC$ (Radii of same circle)

$\implies RC=QC$

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$\implies 5x +x=51+15$

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3 years ago

Which of the following best describes XYZ

Alex787 [66]

<h2>Answer:</h2>

D. Inscribed angle

<h2>Step-by-step explanation:</h2>

An inscribed angle has a vertex on the circumference. For any arc, every inscribed angle is exactly half of the central angle. In other words, if the central angle is $x^{\circ}$ and the inscribed angles is $y^{\circ}$ , then it is true that:

$y^{\circ}=\frac{x^{\circ}}{2}$

In this problem $y^{\circ}=36^{\circ}$ , then the central angle is $x^{\circ}=2y^{\circ} \ \therefore x^{\circ}=2(36^{\circ}) \therefore x^{\circ}=72^{\circ}$