 Hello friends, welcome to the session on Malka. Let's discuss the given question. In figure 6.3a or deduces ad and ce of triangle abc intersect each other at the point p. Show that triangle abc intersect each other at the point p. Show that triangle abdabd is similar to triangle ababp is similar to triangle abdb and 4th triangle pdc is similar to triangle beec. Now let's begin with the solution. We are given that ad is perpendicular to cb, ce is perpendicular to ab. We are also given that the two attributes ad and ce intersect at point p. We have to show is similar to triangle cdp in triangle a, e, p and triangle cdp triangle cdp. We have angle ap equal to angle cdp equal to 90 degree. Since we are given that ce is perpendicular to ab and ad is perpendicular to cb. Therefore, angle aep equal to angle cdp equal to 90 degree, angle ap equal to angle cpd because they are vertically opposite angle angle ap equal to angle cpd because they are vertically opposite. Now therefore by a criteria of similarity we have triangle aep is similar to triangle cdp. We all know that in a criteria if two angles of one triangle are equal to two angles of another triangle then the two triangles are similar and here also in triangle aep and triangle cdp we have two angles of one triangle are equal to other two triangles of another triangle. So we can say the two triangles are similar ap similar to triangle cdp. So this is proof. Now let's see the next part. We have to show that triangle a, b, b is similar to triangle c, b, e. Let's start with the solution. In triangle a, b, b and triangle c, b, e we have angle a, b, b equal to angle c, b, e, angle a, b, b is equal to c, b, e since b is the common angle. Angle a, b, b equal to angle c, e, b equal to 90 degree, angle a, b, b equal to angle c, e, b equal to 90 degree since we are given that c, e are perpendicular to ab, ad are perpendicular to c, b. Therefore criteria of similarity the two triangles are similar. Triangle a, b, b is similar to triangle c, b and spruce. Hope you understood it. Now let's see the next part. We have to show that triangle a, e, p is similar to triangle a, d, b. Now let's start with the solution. In triangle a, e, p, triangle a, b, b we have angle a, e, p equal to angle a, d, b equal to 90 degree. We can see from the figure that angle a, e, p equal to angle a, d, b equal to 90 degree since ad is perpendicular to b, c and ce is perpendicular to ab. Now angle p, a, e equal to angle d, a, b, common angle as we can see from the figure that p, a, e is equal to d, a, b since a is the common angle. Therefore by a, a criteria of similarity we have triangle a, e, p similar to triangle a, d, b. So this is proved. Hope you understood it. Now let's see the next part. We have to show that triangle p, d, c is similar to triangle b, e, c. Now let's start with the solution. In triangle p, d, c and triangle b, e, c we have angle p, d, c equal to angle b, e, c equal to 90 degree. We can see from the figure that angle p, d, c equal to b, e, c equal to 90 degree since we have given that ce is perpendicular to ab and ad is perpendicular to cb. Also angle p, c, d equal to angle e, c, b, common angle we can see from the figure angle p, c, b equal to angle e, c, b, c is the common angle. Therefore by criteria similarity we have triangle p, v, c similar to triangle b, e, c. Hence by angle p, d, c is similar to triangle b, e, c which was the required result. Hope you understood the solution of all the four parts and enjoyed them. Goodbye and take care. .