Answer:
By exterior angle theorem, we have;
∠DBE = ∠H + ∠HEB = ∠ECD = ∠H + ∠HDC
∴ ∠H + ∠HEB = ∠H + ∠HDC
By addition property of equality, we have
∠HEB = ∠HDC
∠H = ∠H by reflexive property
∴ ΔHCD ~ ΔHEB by Angle Angle AA similarity postulate
∴ HE/HD = EB/DC, by the definition of similarity
Therefore, by cross multiplication, we have;
HE × DC = EB × HD
Therefore, by commutative property of multiplication, we have;
HE × DC = HD × EB
Explanation: