In given figure- If BD AC and CE AB- then prove thati AEC- ADBii CA AB - CE DB

(i) In 'sAEC and ADB , we have #160; #160; #160; #160; ∠ AEC=∠ ADB=90^0 #160; #160; #160; #160;[ ∵ CE AB and BD ...

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Byju's Answer

Standard VI

Mathematics

Quadrilaterals

In given figu...

Question

In given figure, If $B D ⊥ A C$ and $C E ⊥ A B$ , then prove that

(i) $△ A E C \sim △ A D B$

(ii) $\frac{C A}{A B} = \frac{C E}{D B}$

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Solution

(i) In $△^{'} s A E C$ and $A D B$ , we have

$∠ A E C = ∠ A D B = 90^{0}$ [ $∵ C E ⊥ A B$ and $B D ⊥ A C$ ]

and, $∠ E A C = ∠ D A B$ [Each equal to $∠ A$ ]

Therefore, by AA-criterion of similarity, we have

$△ A E C \sim △ A D B$ $[H e n c e p r o v e d]$

(ii) we have,

$△ A E C \sim △ A D B$ [AS proved above]

$\Rightarrow$ $\frac{C A}{B A} = \frac{E C}{D B}$

$\Rightarrow$ $\frac{C A}{A B} = \frac{C E}{D B}$ $[H e n c e p r o v e d]$

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In given figure, If BD⊥AC and CE⊥AB, then prove that(i) △AEC∼△ADB(ii) CAAB=CEDB

(i) In △′sAEC and ADB, we have ∠AEC=∠ADB=900 [∵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]⇒ CABA=ECDB⇒ CAAB=CEDB [Hence proved]

In given figure, If $B D ⊥ A C$ and $C E ⊥ A B$ , then prove that

(i) $△ A E C \sim △ A D B$

(ii) $\frac{C A}{A B} = \frac{C E}{D B}$