The following figure shows a triangle ABC in which AD and BE are perpendiculars to BC and AC respectively.

Show that :

(i) △ADC ~ △BEC

(ii) CA × CE = CB × CD

(iii) △ABC ~ △DEC

(iv) CD × AB = CA × DE

(i) In △ADC and △BEC,

∠ADC = ∠BEC [Both = 90°]

∠ACD = ∠BCE [Common angle]

∴ △ADC ~ △BEC [By AA].

Hence, proved that △ADC ~ △BEC.

(ii) Since, △ADC ~ △BEC.

We know that,

Corresponding sides of similar triangles are equal.

$\therefore \dfrac{CA}{CB} = \dfrac{CD}{CE}$

⇒ CA × CE = CB × CD.

Hence, proved that CA × CE = CB × CD.

(iii) From part (ii) we get,

$\Rightarrow \dfrac{CA}{CB} = \dfrac{CD}{CE} \\[1em] \Rightarrow \dfrac{CA}{CD} = \dfrac{CB}{CE}$

∠DCE = ∠BCA [Common angle]

∴ △ABC ~ △DEC [By SAS]

Hence, proved that △ABC ~ △DEC.

(iv) Since, △ABC ~ △DEC

We know that,

Corresponding sides of similar triangles are proportional to each other.

⇒ $\dfrac{CA}{CD} = \dfrac{AB}{DE}$

⇒ CD × AB = CA × DE.

Hence, proved that CD × AB = CA × DE.