Question

In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P.Show that:

(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ Δ ADB
(iv) ΔPDC ~ ΔBEC

Answer 1

(i)
In ∆AEP and ∆CDP,
$\angle$AEP = $\angle$CDP (Each 90°)
$\angle$APE = $\angle$CPD (Vertically opposite angles)
Hence, by using AA similarity criterion,
∆AEP ∼ ∆CDP

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(ii)
In ∆ABD and ∆CBE,
$\angle$ADB = $\angle$CEB (Each 90°)
$\angle$ABD = $\angle$CBE (Common)
Hence, by using AA similarity criterion,
∆ABD ∼ ∆CBE

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(iii)
In ∆AEP and ∆ADB,
$\angle$AEP = $\angle$ADB (Each 90°)
$\angle$PAE = $\angle$DAB (Common)
Hence, by using the AA similarity criterion,
∆AEP ∼ ∆ADB

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(iv)
In ∆PDC and ∆BEC,
$\angle$PDC = $\angle$BEC (Each 90°)
$\angle$PCD = $\angle$BCE (Common angle)
Hence, by using the AA similarity criterion,
∆PDC ∼ ∆BEC