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.

## Problem

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.

## Strategy

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°.

Finally, from the two tangent theorem, we know that AB=AC, so ΔABC is an isosceles triangle, and by the base angle theorem, m∠ABC=m∠ACB, so they each must be 45°.

## Solution

(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