Question

Given: In the figure, point A is in the exterior of the circle with centre P. AB is the tangent segment and secant through A intersects the circle in C and D.

To prove: AB² = AC × AD

Construction: Draw segments BC and BD.

Write the proof by completing the activity.

Proof: In ΔABC and ΔADB,

∠BAC ≅ ∠DAB  .....becuase ______

∠______ ≅ ∠______  ......[Theorem of tangent secant]

∴ ΔABC ∼ ΔADB  .......By ______ test

∴ `square/square = square/square`   .....[C.S.S.T.]

∴  AB² = AC × AD

Proved.

Answer 1

Proof: In ΔABC and ΔADB,

∠BAC ≅ ∠DAB  .....because common angle

∠ABC ≅ ∠ADB  ......[Theorem of tangent secant]

∴ ΔABC ∼ ΔADB  .......By AA test

∴ `bb(AB)/bb(AD) = bb(AC)/bb(AB)`   .....[C.S.S.T.]

∴  AB² = AC × AD

Proved.