In adjoining figure, BD perpendicular to AC and CE perpendicular to AB then prove that
(a) ΔAEC similar to ΔADB
(b)CA/AB=CE/DB
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Step-by-step explanation:
(i) In △
′
sAEC and ADB, we have
∠AEC=∠ADB=90
0
[∵CE⊥AB and BD⊥AC]
and, ∠EAC=∠DAB [Each equal to ∠A]
Therefore, by AA-criterion of similarity, we have
△AEC∼△ADB [Hence proved]
(ii) we have,
△AEC∼△ADB [AS proved above]
⇒
BA
CA
=
DB
EC
⇒
AB
CA
=
DB
CE
[Hence proved]
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