Consider a square ABCD. Let L be a point on the diagonal AC. Consider two squares APLQ and CMLN

Consider a square ABCD. Let L be a point on the diagonal AC. Consider two squares APLQ and CMLN, contained in the original square, with a common vertex L, where point P lies on side AB. Let O be the center of the second square CMLN. Find the corner of the PDO.

Let’s denote the side of the square APLQ as a, the side of the square CMLN as b;

Let us drop the perpendiculars from point O to the sides LN and CN and get points E and F;

Triangles OPE and ODF are equal on two sides and a right angle between them: OE = OF = b / 2; EP = DF = a + b / 2;

Then the angle ∠OPE = ∠ODF and PO = OD;

The POD triangle is isosceles and the OPD and ODP angles are equal;

In a right-angled triangle PND ∠NPD + ∠NDP = 180 – 90 = 90 °;

Since ∠OPE = ∠ODF the sum ∠OPD + ∠ODP = ∠OPE + ∠NPD + ∠NDP – ∠ODF = ∠NPD + ∠NDP = 90 °;

Since the angles OPD and ODP are equal, ∠ODP = 90 ° / 2 = 45 °;