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

Answer

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Siddharta7

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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!

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse

the answer is c.

step-by-step explanation: