Question

In Fig6.38, 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

Answer:

In △AEP and △CDP,

∠APE=∠CPD (Vertically opposite angle)

∠AEP=∠CDP=90

o

∴ By AA criterion of similarity, △AEP ∼ △CDP

In △ABD and △CBE

∠ADB=∠CEB=90

o

∠B is common

∴ By AA criterion of similarity, △ABD ∼ △CBE

In △AEP and △ADB

∠AEP=∠ADB=90

o

∠A is common

∴ By AA criterion of similarity, △AEP ∼ △ADB

△PDC and △BEC

∠PDC=∠BEC=90

o

∠C is common

∴ By AA criterion of similarity, △PDC ∼ △BEC

Step-by-step explanation:

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Answer 2

Answer:

i) To prove : ∆ AEP ~ ∆ CDP

In ∆ AEP and ∆ CDP,

angle AEP= angle CDP ( Both angles are right angle)

angle APE= angle CPD (Vertically opposite angles )

∆ AEP ~ ∆ CDP ( By AA criterion)

ii) To prove : ∆ ABD ~ ∆ CBE

angle ADB = angle CEB (each 90°)

angle ABD = angle CBE. (common)

∆ ABD ~ ∆ CBE. (AA)

iii) to prove : ∆ AEP ~ ∆ ADB

angle AEP = angle ADB. (each 90°)

angle A = angle A. (common)

∆ AEP ~ ∆ ADB. (AA)

iv) to prove : ∆ PDC ~ ∆ BEC

angle PDC = angle BEC. (each 90°)

angle PCD = angle BCE. (common)

∆ PDC = ∆ BEC. (AA)

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