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vichka [17]
3 years ago
7

PLEASE HELP!! AOC and BOD are diameters of a circle, centre O. Prove that triangle ABD and triangle DCA are congruent by RHS.

Mathematics
2 answers:
goldfiish [28.3K]3 years ago
4 0
Give the person their brainliest
Alona [7]3 years ago
4 0

Answer

3.7/5

6

Siddharta7

Genius

2.1K answers

230.1K people helped

Step-by-step explanation:

Given,

BAD and ADC are right angles[angles in semicircle is 90]

AD is common to both triangles.

=> ∠BAD = ∠CDA[angles in semicircle is 90°]

=> ∠ABO = ∠DCA[angles in same segment are equal]

∴ ΔABD and ΔDCA are congruent. ----- ASA

(or)

BD = CA[diameters of the circle]

=> ∠BAD = ∠CDA[angles in semicircle is 90°]

AD is common to both triangles.

ΔABD and ΔDCA are congruent triangles. ----- RHS

(or)

BD = CA[diameter of circle]

AD is common.

=> ∠ADB = ∠CAD[base angles of an isosceles triangle are equal]

∴ ΔABD and ΔDCA are congruent triangles. ------ SAS

Hope it helps!

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lilavasa [31]

Answer:

see explanation

Step-by-step explanation:

Given the 3 equations

3x + 5y + 5z = 1 → (1)

x - 2y = 5 → (2)

2x + 4y = 11 → (3)

Use (2) and (3) to solve for x and y

Multiply (2) by 2

2x - 4y = 10 → (4)

Add (3) and (4) term by term

4x = 21 ( divide both sides by 4 )

x = \frac{21}{4\\}

Substitute this value of x into (3)

2 × \frac{21}{4\\} + 4y = 11

\frac{21}{2\\} + 4y = 11 ( subtract \frac{21}{2\\} from both sides )

4y = \frac{1}{2} ( divide both sides by 4 )

y = \frac{1}{8\\}

Substitute the values of x and y into (1) and solve for z

3 × \frac{21}{4\\} + 5 × \frac{1}{8\\} + 5z = 1

\frac{63}{4} + \frac{5}{8} + 5z = 1

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5z = - \frac{123}{8} ( divide both sides by 5 )

z = - \frac{123}{40}

Solution is

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3 years ago
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