婆罗摩笈多定理

证明

• 因为 ∠ AMF=∠ EMC=∠ MBC=∠ MAD
• 所以 AF=MF
• ∠ ∠ FMD=90^(∘)-∠ AMF=90^(∘)-∠ MAD=∠ FDM
• 因此 MF=DF
• AF=MF=DF
• 即 EF 平分 AD

婆罗摩笈多四边形

婆罗摩笈多四边形是四边形，其边长顺序分别为 a₁b₃,a₃b₂,a₂b₃,a₃b₁，且 a₁²+a₂²=a₃²、 b₁²+b₂²=b₃²。

婆罗摩笈多四边形既是圆内接四边形，又是正交四边形（对角线垂直的四边形）。

In geometry, Brahmagupta's theorem states that if a cyclic quadrilateral is orthodiagonal (that is, has diagonals that are perpendicular), then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side. It is named after the Indian mathematician Brahmagupta (598-668).

More specifically, let A, B, C and D be four points on a circle such that the lines AC and BD are perpendicular. Denote the intersection of AC and BD by M. Drop the perpendicular from M to the line BC, calling the intersection E. Let F be the intersection of the line EM and the edge AD. Then, the theorem states that F is the midpoint of AD.

Proof

We need to prove that AF = FD. We will prove that both AF and FD are in fact equal to FM.

To prove that AF = FM, first note that the angles ∠FAM and ∠CBM are equal, because they are inscribed angles that intercept the same arc of the circle (CD). Furthermore, the angles ∠CBM and ∠CME are both complementary to angle ∠BCM (i.e., they add up to 90°), and are therefore equal. Finally, the angles ∠CME and ∠FMA are the same. Hence, △AFM is an isosceles triangle, and thus the sides AF and FM are equal.

The proof that FD = FM goes similarly: the angles ∠FDM, ∠BCM, ∠BME and ∠DMF are all equal, so △DFM is an isosceles triangle, so FD = FM. It follows that AF = FD, as the theorem claims.