In ΔABC, ∠B = 90° and P is the mid-point of AC. Prove that PB = PA =   AC.

In the ΔABC, P is the mid-point of AC and ∠B = 90°. 
To Prove: PA = PB = AC
Construction: Draw PK ïï BC
Proof: Since KP ïï BC with transversal AB
.^.. ∠1 = ∠B = 90°
Also, ∠1 + ∠2 = 180°
.^.. ∠2 = 180° - ∠1 = 180° - 90° = 90°
Now in ΔAPK and ΔBPK,
KP = KP...(common)
∠1 = ∠2...(prove above)
AK = KB...(since KP ïï BC and P is the mid-point of AC)
.^.. ΔAPK ≅ ΔBPK
.^.. PA = PB  but
PA = AC ...( P is the mid-point of AC)
PA = PB = AC.