Question

In △OAB,∠AOB=90∘. Let C be the point on the segment AB such that OC⊥−AB.

Prove

|CA|/|CB| = |OA|^2/|OB|^2

Answer 1

Answer on Question #82224 - Math - Geometry

Question

1. In $\Delta$ OAB, $\angle AOB = 90{}^{\circ}$ . Let C be the point on the segment AB such that OC⊥-AB.

Prove

$$| C A | / | C B | = | O A | ^ {2} / | O B | ^ {2}$$

Solution

According to the triangles similarity theorem: If two angles of one triangle are equal to two angles of another, then the triangles are similar. $\Delta$ OAB, $\Delta$ OAC, $\Delta$ OBC - similar.

$\angle AOB = \angle OCA = \angle OCB = 90{}^{\circ}$ , $\angle OBA = \angle OBC = \angle AOC = \alpha$ and $\angle OAB = \angle OAC = \angle COB = 90{}^{\circ} - \alpha$ .

So can write:

$$\tan \alpha = \frac {| O A |}{| O B |} = \frac {| C A |}{| O C |} = \frac {| O C |}{| C B |} (1),$$ $$\sin^ {2} \alpha + \cos^ {2} \alpha = 1 (2),$$ $$\frac {| C A | ^ {2}}{| O A | ^ {2}} + \frac {| C B | ^ {2}}{| O B | ^ {2}} = 1,$$ $$\frac {| C A | ^ {2}}{| O A | ^ {2}} = 1 - \frac {| C B | ^ {2}}{| O B | ^ {2}} = \frac {| O B | ^ {2} - | C B | ^ {2}}{| O B | ^ {2}},$$ $$\frac {| C A | ^ {2}}{| O B | ^ {2} - | C B | ^ {2}} = \frac {| O A | ^ {2}}{| O B | ^ {2}},$$

where $|OB|^2 - |CB|^2 = |OC|^2$ , so have

$$\frac {| C A | ^ {2}}{| O C | ^ {2}} = \frac {| O A | ^ {2}}{| O B | ^ {2}},$$

from (1) obtain $|OC|^2 = |CA| \cdot |CB|$ .

As a result,

$$\frac {| \mathrm {C A} | ^ {2}}{| \mathrm {C A} | \cdot | \mathrm {C B} |} = \frac {| \mathrm {C A} |}{| \mathrm {C B} |} = \frac {| \mathrm {O A} | ^ {2}}{| \mathrm {O B} | ^ {2}}$$

Answer provided by https://www.AssignmentExpert.com