Question

Identify the three similar right triangles in the given diagram.

Answer 1

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.