# In the figure, altitudes AD and CE of $$\triangle$$ ABC intersect each other at the point P. Show that : (i) $$\triangle$$ AEP ~ $$\triangle$$ CDP (ii) $$\triangle$$ ABD ~ $$\triangle$$ CBE (iii) $$\triangle$$ AEP ~ $$\triangle$$ ADB (iv) $$\triangle$$ PDC ~ $$\triangle$$ BEC

## Solution :

(i)  In $$\triangle$$ AEP and CDP, we have

$$\angle$$ AEP = $$\angle$$ CDP = 90

$$\angle$$ APE = $$\angle$$ CPD           (vertically opposite angles)

$$\therefore$$  By AA similarity, we have :

$$\triangle$$ AEP ~ $$\triangle$$ CDP

(ii)  In $$\triangle$$ ABD and CBE, we have

$$\angle$$ ABD = $$\angle$$ CBE         (common angle)

$$\angle$$ ADB = $$\angle$$ CEB = 90

$$\therefore$$  By AA similarity, we have :

$$\triangle$$ ABD ~ $$\triangle$$ CBE

(iii)  In $$\triangle$$ AEP and ADB, we have

$$\angle$$ AEP = $$\angle$$ ADB = 90

$$\angle$$ PAE = $$\angle$$  DAB          (common angles)

$$\therefore$$  By AA similarity, we have :

$$\triangle$$ AEP ~ $$\triangle$$ ADB

(iv)  In $$\triangle$$ PDC and BEC, we have

$$\angle$$ PDC = $$\angle$$ BEC = 90

$$\angle$$ PCD = $$\angle$$ ECB           (common angles)

$$\therefore$$  By AA similarity, we have :

$$\triangle$$ PDC ~ $$\triangle$$ BEC