Consider two right-angled triangles inscribed in a square such that their hypotenuse intersect at a point O inside the square. What is the angle between the hypotenuse of the triangle and the base of the square?
45∘
In the square ABCD, we have right-angled triangles, △ ADC and △ BCD such that their hypotenuse intersect at the point O inside the square.
Now, since the diagonals of a square are equal in length, and every angle is a right angle, we have
BD=AC and ∠B=90∘.
Thus, △BDC is a right-angled isosceles triangle.
⟹∠BDC=45∘ .....(Sum of angles of a triangle is 180∘)