In the given figure, altitudes AD and CE of ∆ ABC intersect each other at the point P. Show that:

(i) ∆ AEP ~ ∆ CDP

(ii) ∆ ABD ~ ∆ CBE

(iii) ∆ AEP ~ ∆ ADB

(iv) ∆ PDC ~ ∆ BEC

(i) In Δ AEP and Δ CDP,

⇒ ∠AEP = ∠CDP (Both = 90°)

⇒ ∠APE = ∠CPD (Vertically opposite angles are equal)

⇒ Δ AEP ~ Δ CDP (By A.A. axiom)

Hence, proved that Δ AEP ~ Δ CDP.

(ii) In Δ ABD and Δ CBE,

⇒ ∠ADB = ∠CEB (Both = 90°)

⇒ ∠ABD = ∠CBE (Common angle)

⇒ Δ ABD ~ Δ CBE (By A.A. axiom)

Hence, proved that Δ ABD ~ Δ CBE.

(iii) In Δ AEP and Δ ADB,

⇒ ∠AEP = ∠ADB (Both = 90°)

⇒ ∠PAE = ∠BAD (Common angle)

⇒ Δ AEP ~ Δ ADB (By A.A. axiom)

Hence, proved that Δ AEP ~ Δ ADB.

(iv) In Δ PDC and Δ BEC,

⇒ ∠PDC = ∠BEC (Both = 90°)

⇒ ∠PCD = ∠BCE (Common angle)

⇒ Δ PDC ~ Δ BEC (By A.A. axiom)

Hence, proved that Δ PDC ~ Δ BEC.