 Hi and welcome to the session my name is Shashi and I am going to help you with the following question. Question is in figure 6.57, D is a point on hypotenuse AC of triangle ABC, Dm is perpendicular to BC and Dn is perpendicular to AB, prove that Dm square is equal to Dn multiplied by mc. Second part we have to prove is Dn square is equal to Dm multiplied by an. Let us now start with the solution. First of all we will write whatever is given in the question. Now we are given D is a point on hypotenuse AC of triangle ABC. Also Dn is perpendicular to BC and Dn is perpendicular to AB. Now let us write to prove Dn square is equal to Dn multiplied by mc and second part we have to prove is Dn square is equal to Dm multiplied by an. Now let us do one construction. We will construct BD perpendicular to AC, write construction draw BD perpendicular to AC. Let us start the proof now. Now AC is the hypotenuse in triangle ABC this implies angle B is a right angle. So we can say AB is perpendicular to BC since angle B is equal to 90 degree. So we will write AB is perpendicular to BC cos angle B is equal to 90 degrees. Also we have given Dm is perpendicular to BC this implies AB is parallel to Dm. This is because perpendicular on the same line are parallel to each other. Dm and AB are perpendicular on same line BC. So this implies AB is parallel to Dm. Now let us name this expression as 1. Now we can see CB is perpendicular to AB and Dn is perpendicular to AB, perpendicular to AB. This is given in the question itself. So this implies CB is parallel to Dn as the perpendicular drawn on the same line are parallel to each other. So Dn and CB both are perpendicular on AB. So they are parallel to each other. Let us name this expression as 2. Now from 1 and 2 we get Dm Dn is a right angle since both pairs of opposite sides are parallel and all angles are of 90 degree we have quadrilateral Dm Dn. So we can write from 1 and 2 we get Dm is a Dn is a rectangle implies Dm is equal to Dn. Opposite sides of rectangle are equal. Now in the given figure let us name these angles as 1, 3. That is we have named angle MBD as angle 1, angle BDM as angle 2 and angle CDM as angle 3. Now in triangle BMD angle 1 plus angle 2 plus angle BMD is equal to 180 degree. There is some property of triangle. You know angle BMD is equal to 90 degree since Dm is parallel to BC. So we will substitute for BMD 90 degrees so we get angle 1 plus angle 2 plus 90 degrees is equal to 180 degree. This implies angle 1 plus angle 2 is equal to 90 degrees. Now let us name this expression as 3 and this expression as 4. Now clearly we can see in triangle DNC also angle 3 plus angle C is equal to 90 degree since angle DNC is equal to 90 degree. So by angle some property we can write angle 3 plus angle C is equal to 90 degree. So similarly in triangle DNC angle 3 plus angle C is equal to 90 degrees. Let us name this expression as 5. We know BD is perpendicular to AC by construction so we can write as BD is perpendicular to AC. This implies angle 2 plus angle 3 is equal to 90 degrees. Let us name this expression as 6. Now from expression 4 and expression 6 we get angle 1 is equal to angle 3. So we can write from and 6 we get angle 1 is equal to angle 3. Similarly from 5 and 6 we get angle C is equal to angle 2 is equal to angle 2. We get angle 1 is equal to angle 3 and angle 2 is equal to angle C. So by AA similarity rule we get triangle BMD is similar to triangle DNC. BMD is similar to triangle DMC. So we can write BMD and triangle DNC is equal to angle 3 and angle C is equal to angle 2. So we get triangle BMD is similar to triangle DMC by AA similarity rule. We know these two triangles are similar to each other so the ratio of their corresponding sides are equal so we can write. Therefore BM upon DM is equal to MD upon MC. This implies DN upon DM is equal to DM upon MC. As we know BM is equal to DN as we have proved in expression 3. So here we have substituted for BM DM. This implies DM square is equal to DN multiplied by MC. This completes the first part of the question as we have proved that triangle BMD is similar to triangle DMC. Similarly we can prove triangle BMD is similar to triangle DNA. So we can write similarly triangle BMD is similar to triangle DNA as these two triangles are similar. The ratios of their corresponding sides are equal so we can write. DM upon DN is equal to MD upon NA. Therefore we can write DN upon DN is equal to ND upon NA also below that BMDN is a rectangle. So DM is equal to DN since these are the opposite sides of the rectangle. So we can write DM is equal to BM. Let us name this expression as 7 and this expression as 8. So we can write substituting BM is equal to DM from N7. We get DM upon DN is equal to DN upon NA or we can say this implies DN square is equal to DM multiplied by NA. Or we can write DN square is equal to DM multiplied by N. This is the required proof for the second part and this is the required proof for the first part. This completes the session. Hope you enjoyed the session. Good bye.