Identify the three similar right triangles in the given diagram.A. ABC, BDA, BCDB.ABD, ADC, DBCC. ABD, ACD, DCBD.ADB, ACD, CDB

Answer:B. ΔABD, ΔADC, ΔDBCStep-by-step explanation Step -1 In ΔABD and ΔADC (from figure).∠DAB=∠CAD (common in both triangles) ,∠DBA=∠CDA =90 degree, and ∠BDA=∠DCA (rest angle of the two triangles).therefore ΔABD similar to ΔADC (by AAA similarity theorem). Step -2 In ΔDBC and ΔADC (from figure).∠DCB=∠ACD (common in both triangles) ,∠DBC=∠ADC =90 degree, and ∠CDB=∠CAD (rest angle of the two triangles).therefore ΔDBC similar to ΔADC (by AAA similarity theorem). Step -3 In ΔABD and ΔDBC (from figure).∠BDA=∠BCD (because , ∠ACD=ADB from stap-1 and ∠ACD=∠BCD from figure) ,∠DBA=∠CBD =90 degree, and ∠BAC=∠BDC (rest angle of the two triangles).therefore ΔABD similar to ΔDBC (by AAA similarity theorem).In the above step- ΔABD similar to ΔADC, ΔDBC similar to ΔADC and ΔABD similar to ΔDBC.Hence ΔABD, ΔADC, ΔDBC similar to each other in the given figure.