Answer:
B. ΔABD, ΔADC, ΔDBC
Step-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.
