P is the mid-point of the hypotenuse AB of the right-angled triangle ABC. Prove that : AB = 2 CP.

Given: In Δ ABC, ∠C = 90° and P is mid point of AB.

To prove: AB = 2 CP

Construction: Draw line segment PK parallel to BC, which meets AC at point K.

Proof: Since, PK ∥ BC and AC is traversal,

⇒ ∠AKP = ∠ACB = 90° = ∠PKC

Also, as P is mid-point of AB and KP is parallel to CB; PK bisects side AC i.e., AK = KC.

Now, in Δ APK and Δ CPK,

KP = KP (Common)

∠PKA = ∠PKC (both are 90°)

AK = KC (Since, KP ∥ BC and P is mid point of AB)

By SAS congruency criterion,

Δ APK ≅ Δ CPK

By corresponding parts of congruent triangles,

PA = PC

As we know that P is mid point of AB.

⇒ PA = $\dfrac{1}{2}$ AB

So, PA = PC = $\dfrac{1}{2}$ AB

Hence, AB = 2 CP.