Question

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Answer 1

Answer:

Q9. ∠ABD : ∠ACD = 1 : 1 ; Q10. m∠AOB = 40°

Step-by-step explanation:

Q9.

ΔABC is isosceles Δ ⇒ ∠ABC ≅ ∠ACB

ΔBDC is isosceles Δ ⇒ ∠DBC ≅ ∠DCB

∠ABD = ∠ABC - ∠DBC

∠ACD = ∠ACB - ∠DCB

∠ABD ≅ ∠ACD

∠ABD : ∠ACD = 1 : 1

Q10.

ABCD is a parallelogram and its diagonals are perpendicular and intersect at O. ⇒ ΔAOB, ΔCOB, ΔCOD and ΔAOD are right angled Δs. ⇒ m∠OAB = 90° - 50° = 40°

Answer 2

Answer:

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