Answer:
m∠DAB = 90°
Step-by-step explanation:
* Lets revise some facts about the circle
- If two tangent segments drawn from a point outside the circle, then
the two tangent segments are equal in length
- The radius and the tangent perpendicular to each other at the point
of contact
* Lets solve the problem
∵ AB and AD are two tangent segments to circle C at B and D
respectively
∴ AB = AD ⇒ (1)
∵ CD and CD are two radii
∴ AB ⊥ BC and AD ⊥ DC
∴ m∠ABC = m∠ADC = 90°
∵ m∠BDC = 45°
∵ ∠BDC + m∠ADB = m∠ADC
∴ 45° + m∠ADB = 90° ⇒ subtract 45° from both sides
∴ m∠ADB = 45° ⇒ (2)
- In Δ ABD
∵ AB = AD ⇒ proved in (1)
∴ m∠ABD = m∠ADB ⇒ isosceles triangle
∵ m∠ADB = 45° ⇒ proved in (2)
∴ m∠ABD = 45°
- The sum of the measure of the interior angles of a Δ is 180°
∴ m∠DAB + m∠ABD + m∠ADB = 180°
∴ m∠DAB + 45° + 45° = 180° ⇒ simplify
∴ m∠DAB + 90° = 180° ⇒ subtract 90° from both sides
∴ m∠DAB = 90°
