In this geometry problem, we will combine several theorems – the two tangent theorem, the inscribed angle theorem, the sum of angles in a triangle theorem and the sum of angles in a quadrilateral theorem, in order to arrive at the solution.
A triangle ΔBCD is inscribed in a circle such that m∠BCD=75° and m∠CBD=60°. Show that the triangle ΔABC formed by two tangent lines from point A outside the circle to points B and C is a 45-45-90 Right Triangle.
We are given two of the three angles of triangle ΔBCD, so it is easy and usually useful to get the third angle, using the sum of angles in a triangle theorem. We do that to find that m∠BDC=45°.
We’ve also previously said that most problems involving a polygon inscribed in a circle will make use of the inscribed angle theorem, so let’s draw the corresponding central angle to ∠BDC, ∠BOC, and using the inscribed angle theorem know it measures 90°.
Since AB and AC are tangent lines to the circle, they are perpendicular to the radii OB and OC at the points B and C, so in quadrilateral ABOC we have three angles that measure 90° each, and the remaining angle ∠BAC must be 90° as well since the sum of angles in a quadrilateral is 360°.
(1) m∠BCD=75° //Given
(2) m∠CBD=60° //Given
(3) m∠BDC=45° //(1) , (2) , Sum of angles in a triangle
(4) m∠BOC=90° //(3), inscribed angle theorem
(5) m∠ABO=90° //Given, AB is tangent to O
(6) m∠ACO=90° //Given, AC is tangent to O
(7) m∠BAC=90° //(4) ,(5), (6), Sum of angles in a quadrilateral
(8) AB=AC //Two tangent theorem
(9) m∠ABC=m∠ACB //(8), base angle theorem
(10) m∠ABC=m∠ACB=45° //(7),(9), Sum of angles in a triangle