 Hello and welcome to the session, let's work out the following problem. It says if a chord is drawn to the point of contact of tangent to a circle, then prove that the angles which this chord makes with the given tangent are equal respectively to the angles formed in the corresponding alternate segments. Using the above theorem, prove the following. If triangle ABC is isosceles, with AB is equal to AC, prove that the tangent at A to the circle circle of triangle ABC is parallel to BC. Let's now move on to the solution and let's first write what is given to us. We are given that PQ is a tangent to a circle with center O at point A. AB is a chord, the points in the two segments of the circle formed by the chord AB. So we are given that PQ is a tangent at point A to the circle with center O. AB is the chord and C and D are two points in the two segments of formed by the chord AB. Right? Let's now write what we have to prove. We have to prove that angle BAQ is equal to angle ACB that is this and also angle BAP that is this angle is equal to angle ADB that is this angle. Let's now do some construction. Draw diameter AOE and join EB. So this is the diameter AOE and now we have to join EB. Let's now start the proof. Now in triangle AEB angle ABE that is this angle is 90 degrees because it's an angle in semi-circle. Right? Therefore some of these two angles will be 90 degrees as we know that some of three angles in a triangle is 180 degrees and since angle ABE is 90 degrees therefore some of these two angles will be 90 degrees, some of the remaining angles in a triangle. Therefore angle AEB plus angle EAB is equal to 180 minus angle ABE that is 180 minus 90 degrees that is 90 degrees again as EA is perpendicular to PQ as PQ is the tangent and A is the diameter. So AE is perpendicular to PQ therefore angle EA angle EAB that is this plus BAQ that is this is 90 degrees angle EAB plus BAQ which is actually angle EAQ is equal to 90 degrees. Let's name this as 1 and this as 2. Now from 1 and 2 we have angle AEB is equal to angle BAQ. Right? Because in both the equations angle EAB is common and both some gives 90 degrees. So angle AEB is equal to angle BAQ also angle AEB is equal to angle ACB because angles in the same segment these are the angles in the same segment AEB is equal to BAQ and also angle AEB is equal to ACB. So therefore angle ACB is equal to angle BAQ. Now angle BAQ plus angle BAP is 180 degrees that is this plus this is 180 degrees it's a linear pair and also angle ACB plus angle ABB is equal to 180 degrees that is this angle plus this angle is 180 degrees because these are the opposite angles of cyclic quadrilateral. Now we have seen that BAQ plus angle BAP is 180 degrees and also angle ACB plus ADB is 180 degrees. So let's name this as A and this as B from A and B we have angle BAQ plus angle BAP is equal to angle ACB plus angle ADB because both are equal to 180 degrees. Now we have proved above that angle BAQ is equal to angle ACB. Right? So this gets cancelled on both sides so we have angle BAP is equal to angle ADB. Hence we have proved BAQ is equal to angle ACB and angle BAP is equal to angle ADB. Let's now discuss the second part we have to prove using the above theorem that if triangle ABC is isosceles with AB is equal to AC prove that the tangent at A to the circum circle of triangle ABC is parallel to BC that is we have to prove that PQ is parallel to BC. So let's now discuss the second part let's write what is given to us we are given triangle ABC such that AB is equal to AC and PQ is the tangent to the circle at A and let's now write what we have to prove we have to prove that PQ is parallel to BC and to prove that PQ is parallel to BC will prove that the alternate angles are equal and for that we'll prove that angle BAP that is this angle is equal to angle ABC these are the alternate angles. So let's now start the proof now in the above theorem we've proved that if we have a tangent and from a point of contact if we draw a chord then the angles in the corresponding alternate segments are equal so angle is equal to angle ACB that is this angle is equal to this angle right this is by above theorem and also angle ABC that is this angle is equal to angle ACB since AB is equal to AC and we know that angles opposite the equal sides are equal right now we have proved that angle BAP is equal to ACB and ACB is equal to ABC so we have angle BAP is equal to angle ABC that is angle BAP is equal to angle ABC but these are alternate angles so this implies PQ is parallel to BC as we know that if the alternate angles are equal then two lines are parallel hence proved so this completes the question and the session bye for now take care have a good day