 Hello and welcome to the session I am Shashi and I am going to help you with the following question. Question says in figure 6.60 AD is a medium of a triangle ABC and AM is perpendicular to BC prove that AC square is equal to AD square plus BC multiplied by DM plus square of BC upon 2. Second part is AB square is equal to AD square minus BC multiplied by DM plus square of BC upon 2. Third part is AC square plus AB square is equal to 2 AD square plus half multiplied by BC square. This is figure 6.60. First of all let us understand that if we are given a triangle ABC such that angle ABC is greater than 90 degree and AD is perpendicular to BC then AC square is equal to AB square plus BC square plus 2 BC multiplied by BD. This fact we have already learnt in the previous question. Now if in triangle ABC angle ABC is less than 90 degree and AD is perpendicular to BC then AC square is equal to AB square plus BC square minus 2 BC multiplied by BD. We have already learnt this fact in the previous question. Now we will use these facts as our key idea to solve the given question. Let us now start with the solution. In triangle AMD we know AM is perpendicular to BC this is given in the question itself. So we get angle AMD is equal to 90 degree. Now this implies angle ADM is less than 90 degree by angle some property of triangle we get angle ADM is less than 90 degree. So we can write in triangle AMD angle AMD is equal to 90 degree this implies angle ADM is less than 90 degrees. Now we know angle ADM and angle ADC form a linear pair and we also know that angle ADM is less than 90 degree. So this implies angle ADC is greater than 90 degree. Now we know AD is a median on BC and we know median divides the opposite side into two equal parts. So we can write BD is equal to DC is equal to half BC. Now first of all let us consider triangle ADC in triangle ADC we know angle ADC is greater than 90 degree and AM is perpendicular to DC. So by using key idea one we get AC square is equal to AD square plus DC square plus 2 multiplied by DC multiplied by DM. So we can write AC square is equal to AD square plus DC square plus 2 DC multiplied by DM. Now we know DC is equal to half BC. Let us name this expression as 1. So we can substitute half BC for DC in this expression. So we get AC square is equal to AD square plus BC upon 2 square plus 2 multiplied by BC upon 2 multiplied by DM. Now 2 and 2 will get cancelled and we get AC square is equal to AD square plus BC multiplied by DM plus square of BC upon 2. Now this completes the first part of the given question. Let us now name this expression as 2. Now we will consider triangle ADB in triangle ADB. We know angle ADB is less than 90 degree. AM is perpendicular to BD. Now by using key idea 2 we get AB square is equal to AD square plus BD square minus 2 multiplied by BD multiplied by DM. So we can write AB square is equal to AD square plus BD square minus 2 VD multiplied by DM. Now from expression 1 we know BD is equal to half BC. So we can substitute half BC for VD in this expression. So we can write AB square is equal to AD square plus square of BC upon 2 minus 2 multiplied by BC upon 2 multiplied by DM. Now 2 and 2 will cancel each other and we get AB square is equal to AD square minus BC multiplied by DM plus square of BC upon 2. Now this completes the second part of the question. Now let us name this equation as 3. Now adding expressions 2 and 3 we get AC square plus AB square is equal to AD square plus BC multiplied by DM plus square of BC upon 2 plus AD square minus BC multiplied by DM plus square of BC upon 2. Now these 2 terms will get cancelled and we get AC square plus AB square is equal to we know AD square plus AD square is equal to 2 AD square plus and square of BC upon 2 plus square of BC upon 2 is equal to 2 multiplied by square of BC upon 2. Now simplifying further we get AC square plus AB square is equal to 2 AD square plus 2 multiplied by BC square upon 4. Now here we will cancel common factor 2 from numerator and denominator both and finally we get AC square plus AB square is equal to 2 AD square plus BC square upon 2 or we can write it as AC square plus AB square is equal to 2 AD square plus half BC square. This is the third part of the question. So we have proved all the 3 parts of the question. This is the first part. This is the second part and here we have proved the third part. So this completes the session. Hope you understood the solution. Take care and have a nice day.