Question

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

Answer 1

Step-by-step explanation:

Given: △ABC, P is mid point of AB, ∠C=90

∘

To prove: PA=PB=

2

1

AB

Construction: Draw PK∥BC

Since, PK∥BC with transversal AC,

∠1=∠C=90

∘

Also, ∠1+∠2=180

∠2=180−∠1=180−90=90

∘

Now, In △APK and △CPK,

KP=KP (Common)

∠1=∠2=90

∘

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

△APK≅△CPK (SAS rule)

Therefore, PA=PC

PA=

2

1

AB

PA=PC=

2

1

AB

AB=2PC