Question

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​

Answer 1

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]