 Hello and how are you all today? The question says in a triangle if the square on one side is equal to the sum of the squares of the other two sides, prove that the angle opposite to the first side is a right angle. Use the above theorem and prove the following. In triangle ABC, AD is perpendicular on BC and BD is twice CD. We need to prove that twice BC square is equal to twice EC square plus BC square. So first of all let us prove the stated theorem. For that we need to have this diagram. So we are solving the first part of this question. We are given a triangle such that AB square plus BC square is equal to EC square to prove equal to 90 degree. Now for that we have done a little construction. We have constructed triangle PQR right angle at Q such that QR is equal. Now let us start with our proof. Now QR we have equal to 90 degree that is by construction that we have. So by using the chorus theorem we can write that PR square is equal to PQ square plus QR square right. But we have taken in our construction that PQ is equal to AB and QR is equal to BC right. So we can write that PR square is equal to in place of PQ we can have AB square and in place of QR we can have BC square right. Also the square is equal to AB square plus BC square. It is given to us in the question. So these two equations implies that PR square is equal to EC square or we can see that PR is equal to. So congruent to triangle PQR by SA congruency that is all the three sides are equal to each other. PQ is equal to AB, QR is equal to BC and we have proved above that PR is equal to EC. So we can see that therefore angle B is equal to angle Q is equal to 90 degree that is by CPCT. So therefore angle B is equal to 90 degree. Now if angle B is equal to 90 degree then we have proved that opposite to first side is a right triangle if the square of one side is equal to the sum of square of the other two sides right. So this completes the first solution. Now let us proceed with the second part here we are given that in this triangle ABC AD is perpendicular to BC and VD is equal to twice CD. We need to prove that twice AB square is equal to twice AC square plus BC square. Now let us begin with our proof. Now in triangle ABD and in triangle we have that AD square is E square minus VD square. Let this be the first equation and here we can see that AD square is equal to AC square minus DC square that is by Pythagoras theorem we have proved it and we have just rearranged the terms let this be the second equation. So from the first and the second equation we get since the difference are both equal to AD square so we have AB square minus BD square is equal to AC square minus DC square but we are given to us that BD is equal to twice CD so we can say that now will be BD plus CD so 3 CD plus CD will give us in total 4 CD right. So we have minus CD square square we have just taken this term from the left hand side to the right hand side. Now in substituting the values we have AB square equal to AC square minus CD square plus 4 CD the whole square that is AB square minus CD square 9 CD square this BD is equal to twice CD itself that is AB square is equal to D square plus 8 CD square and fourth BC right. So in substituting this here we have B square is equal to AC square plus 8 into 1 by 4 BC now in place of CD we have 1 by 4 BC this can be solved 8 into 1 by 16 BC square now taking two here we have 2 AB square is equal to 2 AC square plus square now this was a required thing to be proved so hence we have proved the given question hope you understood the solution well I enjoyed it to have a nice day.