In the adjoining figure- ABCD is a square and triangle EDC is an equilateral triangle- Prove that AE-BE-

Question

In the adjoining figure- ABCD is a square and triangle  EDC is an equilateral triangle- Prove that AE-BE-

Toppr Admin · Accepted Answer

In -x25B3-ADE -amp- -x25B3-BCEAD-BC -given -x2220-ADE-x2220-BCE-90-x2218-60-x2218-150-x2218-DE-CE -given -x2234-x25B3-ADE-x2245-x25B3-BCE -by SAS congruency rule-AE-BE-CPCT-

In the adjoining figure, $A B C D$ is a square and $△ E D C$ is an equilateral triangle. Prove that
$A E = B E$ .

In $△ A D E$ & $△ B C E$
$A D = B C$ (given )
$∠ A D E = ∠ B C E = 90^{\circ} + 60^{\circ} = 150^{\circ}$
$D E = C E$ (given )
$∴ △ A D E ≅ △ B C E$ (by SAS congruency rule)
$A E = B E (C P C T)$

In the adjoining figure, ABCD is a square and △ EDC is an equilateral triangle. Prove that AE=BE.

In △ADE & △BCEAD=BC (given )∠ADE=∠BCE=90∘+60∘=150∘DE=CE (given )∴△ADE≅△BCE (by SAS congruency rule)AE=BE(CPCT)

In the adjoining figure, $A B C D$ is a square and $△ E D C$ is an equilateral triangle. Prove that
$A E = B E$ .

In $△ A D E$ & $△ B C E$
$A D = B C$ (given )
$∠ A D E = ∠ B C E = 90^{\circ} + 60^{\circ} = 150^{\circ}$
$D E = C E$ (given )
$∴ △ A D E ≅ △ B C E$ (by SAS congruency rule)
$A E = B E (C P C T)$