In circle K shown below, points B, C, D, and E lie on the circle with secants HBD and HCE drawn. Prove: HExDC=HDxEB

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

Step-by-step explanation:

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step-by-step explanation: ?