Question

Identify the three similar right triangles in the given diagram.

A. ABC, BDA, BCD
B.ABD, ADC, DBC
C. ABD, ACD, DCB
D.ADB, ACD, CDB

Answer 1

Based on the diagram shown above, the three similar right triangles are: B. ΔABD, ΔADC, ΔDBC.

The Right Triangles Similarity Theorem states that when the altitude of a triangle is drawn to the hypotenuse of a right angled triangle, then, the two triangles that are formed would be similar to each other, as well as the original triangle.

Based on triangles ABD and ADC, we can logically deduce the following congruent angles;

m∠DAB ≅ m∠CAD

m∠DBA ≅ m∠CDA = 90°

m∠BDA ≅ m∠DCA

ΔABD ~ ΔADC (AAA similarity theorem).

Based on triangles DBC and ADC, we can logically deduce the following congruent angles;

m∠DCB ≅ m∠ACD

m∠DBC ≅ m∠ADC = 90°

m∠CDB ≅ m∠CAD

ΔDBC ~ ΔADC (AAA similarity theorem).

Based on triangles ABD and ΔDBC, we can logically deduce the following congruent angles;

m∠BDA ≅ m∠BCD

m∠DBA ≅ m∠CBD = 90

m∠BAC ≅ m∠BDC

ΔABD ~ ΔDBC (AAA similarity theorem).

Answer 2

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.