AOC and BOD are diameters of a circle, centre O. Prove that triangle ABD and triangle DCA are congruent by RHS
Answers
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!
Given:
AOC and BOD are diameters of a circle and has center O.
To Find:
ΔABD and ΔDCA are congruent by RHS.
Solution:
It is given that AOC and BOD are diameters of a circle.
⇒ BD = CA [diameters of the circle]
⇒ ∠BAD = ∠CDA [angles in semicircle is 90°]
⇒ AD = AD [common in both the triangles]
⇒ ΔABD ≅ ΔDCA [using RHS congruence criteria]
Hence, proved ΔABD ≅ ΔDCA by RHS congruency criteria.