Tangent and Intersected Chord Theorem

This proof will consist of two parts. The first and second equations of the theorem will be proved one at a time.

m∠ 1=1/2mAC

Consider a diameter AF. Since AB is tangent to the circle at A, by the Tangent to Circle Theorem, AF and AB are perpendicular.

In addition, since AF is a diameter, ACF is a semicircle; its measure is therefore 180^(∘). By the Arc Addition Postulate, an equation that relates the measures of AC and FC can be written. mAC + mFC = mACF ⇕ mAC + mFC = 180^(∘) Since ∠ CAF is an inscribed angle, by the Inscribed Angle Theorem, its measure is half the measure of the intercepted arc FC. m∠ CAF = 12mFC ⇕ mFC = 2m∠ CAF Therefore, 2m∠ CAF can be substituted for mFC in the equation mAC+mFC=180^(∘). mAC + mFC = 180^(∘) ⇕ mAC + 2m∠ CAF = 180^(∘) Similarly, by the Angle Addition Postulate, it is evident that the sum of the measures of ∠ 1 and ∠ CAF is 90^(∘). m∠ 1 + m∠ CAF = 90^(∘) Next, multiply the equation above by -2 and add it to the previous equation. mAC + 2m∠ CAF &= 180^(∘) ^+ -2m∠ 1 - 2m∠ CAF &= -180^(∘) [-1em] mAC - 2m∠ 1 &= 0 It is now possible to solve for m∠ 1 using this equation. The first equation of the theorem has been obtained.

m∠ 2=1/2mADC

The second equation of the theorem will now be obtained. To do so, the Arc Addition Postulate and the Angle Addition Postulate can be used to set the following pair of equations. mADF + mFC = mADC & (I) m∠ EAF + m∠ CAF = m∠ 2 & (II) Since ADF is a semicircle, its measure is 180^(∘). Additionally, since FA and AB are perpendicular, ∠ EAF is a right angle. Recall that it has been previously stated that mFC = 2m∠ CAF. This information can be substituted in the above equations. 180^(∘) + 2m∠ CAF = mADC & (I) 90^(∘) + m∠ CAF = m∠ 2 & (II) Next, multiply Equation (II) by -2 and add it to Equation (I). 180^(∘) + 2m∠ CAF &= mADC ^+ -180^(∘) - 2m∠ CAF &= -2m∠ 2 [-1em] 0 &= mADC - 2m∠ 2 Finally, by solving the resulting equation for m∠ 2, the second equation of the theorem is obtained.