 Hi and welcome to the session. Let's work out the following question. The question says in the figure triangle ABC and Triangle DBC are on the same base BC AD and BC intersect at O Prove that area of triangle ABC divided by area of triangle DBC is equal to AO divided by DO So this is the figure given to us So let us start with a solution to this question Let us see what is given to us in the question. Now we are given two triangles that is triangle ABC and Triangle DBC we have triangle ABC and Triangle DBC which were made at the common base BC and AD and BC intersect at a point O Now we have to prove that area of triangle ABC divided by area of Triangle DBC is equal to AO divided by DO Now first of all we need to do some construction here. What we do is we draw AM perpendicular to BC and we draw DN perpendicular to BC Now let us start with our proof. We see that area of Triangle ABC is equal to half of base that is BC into altitude that is AM We call this one now we see that area of Triangle DBC is equal to half of BC into DN Now this is DN that is the altitude for this triangle and we call this two now dividing one by two we get area of triangle ABC divided by area of triangle DBC is equal to half into BC into AM divided by half into BC into DN Now we see that half of BC gets cancelled from numerator and denominator So we have area of triangle ABC divided by area of triangle DBC is equal to AM divided by DN And this we call three now we see that in Triangle AOM and Triangle DON Angle AMO is equal to angle DNO Because Measure of each of them is 90 degree because these are the perpendicular drawn Also angle AOM is equal to angle DON that is this angle is equal to this angle because they are vertically Opposite Angles therefore we can say that Triangle AOM is similar to triangle DON by angle angle similarity criterion Therefore we can say that AM upon DN is equal to AO Divided by DO because they are corresponding size of similar triangles Now we call this for So we see that from three and four we have area of triangle ABC Divided by area of triangle DBC is equal to AM Divided by DN that is equal to AO divided by DO Therefore we can say that area of triangle ABC divided by area of triangle DBC is equal to AO divided by DO So this is what we were supposed to prove in this question I hope that you understood the solution and enjoyed the session. Have a good day