for all the class x aspirants, dear chlildren kindly go through it.?

Question

In a quadrilateral ABCD sum of angles A and D is 90 deg then prove that AC2 + DB2 =AD2 + BC2

Answer 1

The claim is true if we add the condition that the quadrilateral is convex.

If angles BAD and CDA are on the same side of line AD, then we can extend rays AB and DC until the intersect at a new point, E. Since angles A and D sum to 90, E is a right angle.

By several applications of the Pythagorean Theorem, we find:
AC^2 + DB^2 = (AE^2 + EC^2) + (BE^2 + ED^2)
= (AE^2 + ED^2) + (BE^2 + EC^2)
= AD^2 + BC^2,
which is what we wanted to show.

On the other hand, without the requirement that ABCD be a convex polygon, the claim is false. For example, make triangle CDB with DB=1, a right angle at D, and CD=2. Put A inside this triangle, with AD=AB=1. Now triangle ADB is equilateral, so angle BAD measures 60; meanwhile, angle ADC measures 30. So, they are complementary, as required. However, DB=AD and BC>AC, so AC^2 + DB^2 < AD^2 + BC^2.