Answer:
A). AC ≅ BC
Step-by-step explanation:
Given triangle ACB is bisected at C by CP making 90 degree at P making two triangles, Δ ACP and Δ BCP
It is given in the diagram that AP = PB and ∠APC = ∠BPC = 90°
Since, CP is common to both the triangles, Δ ACP and Δ BCP.
Therefore, Δ ACP is similar to Δ BCP.
AP = PB
CP = CP
∠APC = ∠BPC
Therefore Δ ACP is similar to Δ BCP by side, side, angle (SSA)
And since Δ ACP and Δ BCP are similar, therefore AC = BC
Thus, AC ≅ BC
