Question

part a
in right triangle ( abc ), altitude ( cd ) is drawn to its hypotenuse. select all triangles that must be similar to triangle ( abc ).
( square bac )
( square bcd )
( square acd )
( square cbd )

Explanation:

To determine which triangles are similar to right triangle \( ABC \), we use the AA (Angle - Angle) similarity criterion. In right triangle \( ABC \) with right angle \( \angle ACB = 90^{\circ} \) and altitude \( CD \) to hypotenuse \( AB \):

1. For \( \triangle BCD \):

• \( \angle CDB=90^{\circ} \) (since \( CD \perp AB \)) and \( \angle B \) is common to both \( \triangle ABC \) and \( \triangle BCD \). By AA similarity, \( \triangle ABC \sim\triangle BCD \).

1. For \( \triangle ACD \):

• \( \angle CDA = 90^{\circ} \) (since \( CD\perp AB \)) and \( \angle A \) is common to both \( \triangle ABC \) and \( \triangle ACD \). By AA similarity, \( \triangle ABC\sim\triangle ACD \).

1. For \( \triangle BAC \):

• \( \triangle BAC \) is the same as \( \triangle ABC \), so they are congruent (and thus similar) by the reflexive property.

1. For \( \triangle CBD \):

• \( \triangle CBD \) is the same as \( \triangle BCD \) (just named in a different order), so it is also similar to \( \triangle ABC \) for the same reasons as \( \triangle BCD \).

Answer:

A. \( BAC \) (since \( \triangle BAC\) is the same as \( \triangle ABC\), so they are similar)
B. \( BCD \) (by AA similarity: \( \angle CDB = \angle ACB=90^{\circ}\), \( \angle B\) is common)
C. \( ACD \) (by AA similarity: \( \angle CDA=\angle ACB = 90^{\circ}\), \( \angle A\) is common)
D. \( CBD \) (same as \( \triangle BCD\), so similar to \( \triangle ABC\))