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