 Hello and welcome to the session. I am Asha and I am going to help you with the following problem that says, A t is an altitude of an isosceles triangle A p c in which A p is equal to A c show that first A t bisects B c, second A t bisects angle A. Let us now begin with the solution here we are given triangle A b c in which A b is equal to A c and A d is an altitude so the required figure will be figure will be like this where A b is equal to A c and angle A t b and angle A t c are of 90 degree each and we have to prove A t bisects B c since we have to prove first and second we have to prove A t bisects angle A. Let us now begin with the solution in triangle A d b and triangle A d c side A b is equal to A c this is given and in triangle A d c A c is the hypotenuse or in triangle A d b A b is the hypotenuse so A p is equal to A c are the two hypotenuse of these two triangles which are equal also A d is equal to A d this is because their common side so both the triangles A d b is equal to angle A d c is equal to 90 degree each this is also given to us therefore by RHS rule both the triangles will be congruent that is A b t is congruent to triangle A c t this implies B d is equal to D c since by C p c t or we can say that A d bisects B c also angle B A d is equal to angle C A d this is again by C p c t since both these triangles are congruent therefore corresponding angles are also equal or in other words we can say that A d bisects angle A so these are the two parts we have to prove and hence we can say that A d bisects B c and A d bisects angle A so this completes the solution hope you enjoyed it take care and have a good day.