Question 7:
In the following 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
(v) ∆PDC ∼ ∆BEC
Answer:
(i)
In ∆AEP and ∆CDP,
∠AEP = ∠CDP (Each 90°)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by using AA similarity criterion,
∆AEP ∼ ∆CDP
(ii)
In ∆ABD and ∆CBE,
∠ADB = ∠CEB (Each 90°)
∠ABD = ∠CBE (Common)
Hence, by using AA similarity criterion,
∆ABD ∼ ∆CBE
(iii)
In ∆AEP and ∆ADB,
∠AEP = ∠ADB (Each 90°)
∠PAE = ∠DAB (Common)
Hence, by using AA similarity criterion,
∆AEP ∼ ∆ADB
(iv)
In ∆PDC and ∆BEC,
∠PDC = ∠BEC (Each 90°)
∠PCD = ∠BCE (Common angle)
Hence, by using AA similarity criterion,
∆PDC ∼ ∆BEC
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