cx is an altitude in triangle abc. which statements are true? select two options. □△abc≅△bxc □△axc∼△cxb…

cx is an altitude in triangle abc. which statements are true? select two options. □△abc≅△bxc □△axc∼△cxb □△bcx≅△acx □△acb∼△axc □△cxa≅△cba
Answer
Explanation:
Step1: Recall similarity - right - triangle rules
In right - triangle (ABC) with altitude (CX) from right - angle (C) to hypotenuse (AB), we use the geometric mean theorem for right - triangles. Triangles (\triangle AXC) and (\triangle ACB) share angle (A) and both have a right - angle ((\angle AXC=\angle ACB = 90^{\circ})). Also, triangles (\triangle AXC) and (\triangle CXB) are right - triangles. (\angle AXC=\angle CXB = 90^{\circ}), and (\angle A+\angle ACD = 90^{\circ}), (\angle B+\angle ACD=90^{\circ}), so (\angle A=\angle BCD).
Step2: Check similarity of triangles
- For (\triangle AXC) and (\triangle CXB):
- (\angle AXC=\angle CXB = 90^{\circ}).
- (\angle A+\angle ACX = 90^{\circ}) and (\angle B+\angle BCX = 90^{\circ}), and (\angle ACX+\angle BCX=\angle ACB = 90^{\circ}), so (\angle A=\angle BCX) and (\angle B=\angle ACX). By the AA (angle - angle) similarity criterion, (\triangle AXC\sim\triangle CXB).
- For (\triangle ACB) and (\triangle AXC):
- (\angle A) is common to both triangles and (\angle ACB=\angle AXC = 90^{\circ}). By the AA (angle - angle) similarity criterion, (\triangle ACB\sim\triangle AXC).
Answer:
(\triangle AXC\sim\triangle CXB), (\triangle ACB\sim\triangle AXC)