Question

AOC and BOD are diameters of a circle prove that triangle ABD and DCA are congruent by RHS

Answer 1

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

I hope it will help you to solve your question

Answer 2

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

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