In the adjoining figure, △ABC is right- angled at B and P is the midpoint of AC. Show that PA=PC=PB.
I have an exam tomorrow, please explain ASAP.
Dear Student,
Given, ABC is a right angle triangle, having right angled at B and P is mid - point of AC. 
Given, ABC is a right angle triangle, having right angled at B and P is mid - point of AC.
Proof: Given P is mid - point of AC.
Now, to prove,
we have to produce BP to E so that BP = PE. Join EC.
we have to produce BP to E so that BP = PE. Join EC.
In ∆APB and CPE, we have
AP = PC [As P is mid point of AC]
BP = PE [ Construction ]
∠1 = ∠2 [ Vertically opposite angles ]
∴ ∆ APB ≅ ∆ CPE (SAS congruency)
⇒ AB = EC and ∠ CEP = ∠ABP [ C.P.C.T. ] -----(2)
Thus transversal BE cuts AB and CE such that the alternate angles ∠CEP and ∠ABP are equal so,
CE || AB
⇒∠ABC + ∠ECB = 180° [ Sum of interior angles on the same side of transversal BC intersecting parallel lines AB and CE ]
⇒ 90° + ∠ ECB = 180° [ Right angled at B ]
⇒ ∠ ECB = 90°
Now, In ∆ ABC and ECB, we have
AB = EC [ From (2) ]
BC = CB [ Common ]
and ∠ABC = ∠ ECB = 90°
∴ ∆ABC ≅ ∆ECB
⇒ AC = BE
