Question

P is the midpoint of the hypotenuse AB of the right angle triangle ABC prove that AB = 2 CP

Answer 1

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Hope this will help you..................:-)

Answer 2

Proof:

Construction : CP is produced to D such that CD = CD, line DA is drawn

In ΔBPC and ΔDPA

DP = PC   (by construction)

AP = PB   (Given)

∠DPA = ∠BPC

∴ ΔBPC ≅ ΔDPA

∴ DA = BC

And ∠PDA = ∠PCB

∴ DA ║ BC

∴ ∠DAC + ∠BCA = 180°

∴ ∠DAC = 180° - ∠BCA = 180° - 90° = 90°

In ΔDAC and ΔBCA

DA = BC

AC is common

∠DAC = ∠BCA = 90°

∴ ΔDAC ≅ ΔBCA

∴ DC = AB

or, DC/2 = AB/2

or, CP = AB/2

or, AB = 2CP          (Proved)

Hope this helps.