 Hello and welcome to the session. In this session we discussed the following question which says, in the given figure the point D divides the side PC of triangle ABC in the ratio m is to n proves that area of triangle ABD is to area of triangle ADC is equal to m is to n. Before moving on to the solution let's recall the formula for the area of a triangle and this is equal to half the product of its base and the corresponding altitude. That is area of triangle is equal to half into base into altitude. This is the key idea for this question. Now let's move on to the solution. This is the figure given to us. In this we have that this point D divides the side BC of triangle ABC in the ratio m is to n that is BD is to DC is equal to m is to n and we need to prove that area of the triangle ABD is to the area of the triangle ADC is equal to m is to n. First of all let's calculate the area of the triangle ABD this is equal to half into its base that is BD into its altitude that is AL. Now the area of the triangle ADC is equal to half into its base that is DC into its altitude that is AL. Now area of triangle ABD upon the area of the triangle ADC is equal to half into BD into AL upon half into DC into AL. Now here AL AL cancels half half cancels so we left with area of triangle ABD upon the area of the triangle ADC is equal to BD upon DC. Now we are given that BD is to DC is equal to m is to n so this is equal to m upon n since BD is to DC is equal to m is to n. So we have now got area of triangle ABD upon area of triangle ADC is equal to m upon n that is area of triangle ABD is to area of the triangle ADC is equal to m is to n. We were supposed to prove this so we have proved this hence proved. So this completes the session hope you have understood the solution for this question.