Question

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

Answer 1

$\huge\underbrace{\purple{A}\orange{N}\red{S}\pink{W}\blue{eE}{R}\blue{.}\green{.}}$

solution

Given,

that AD and CE are the altitudes of triangle ABC and these altitudes intersect each other at P.

━━━━━━━━━//━━━━━━━━

(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (90° each)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by AA similarity criterion,

ΔAEP ~ ΔCDP

━━━━━━━━━//━━━━━━━━

━━━━━━━━━//━━━━━━━━

(ii) In ΔABD and ΔCBE,

∠ADB = ∠CEB ( 90° each)

∠ABD = ∠CBE (Common Angles)

Hence, by AA similarity criterion,

ΔABD ~ ΔCBE

━━━━━━━━━//━━━━━━━━

━━━━━━━━━//━━━━━━━━

(iii) In ΔAEP and ΔADB,

∠AEP = ∠ADB (90° each)

∠PAE = ∠DAB (Common Angles)

Hence, by AA similarity criterion,

ΔAEP ~ ΔADB

━━━━━━━━━//━━━━━━━━

━━━━━━━━━//━━━━━━━━

(iv) In ΔPDC and ΔBEC,

∠PDC = ∠BEC (90° each)

∠PCD = ∠BCE (Common angles)

Hence, by AA similarity criterion,

ΔPDC ~ ΔBEC

━━━━━━━━━//━━━━━━━━

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