ABDF is a square and BC - EF in the given figure- Prove that i- Delta ABC cong Delta AFEii- Delta ACG cong Delta AEG

Question

Toppr Admin · Accepted Answer

Given-BC-EFSince ABCD is a square- sides are equalAF-AB-x2220-B-x2220-FTherefore from SAS theoremABC-x2245-AFE-x27F9-AC-AE Hence the triangle is isoceles-xA0-AEG-x2245-ACGTherefore from ASA theoremACG-x2245-AEG

$A B D F$ is a square and $B C = E F$ in the given figure. Prove that
i) $Δ A B C ≅ Δ A F E$
ii) $Δ A C G ≅ Δ A E G$

Given,

$B C = E F$

Since ABCD is a square, sides are equal

$A F = A B$

$∠ B = ∠ F$

Therefore from SAS theorem

$A B C ≅ A F E$

$⟹ A C = A E$ Hence the triangle is isoceles.
$A E G ≅ A C G$

Therefore from ASA theorem

$A C G ≅ A E G$

ABDF is a square and BC=EF in the given figure. Prove that i) ΔABC≅ΔAFEii) ΔACG≅ΔAEG

Given,BC=EFSince ABCD is a square, sides are equalAF=AB∠B=∠FTherefore from SAS theoremABC≅AFE⟹AC=AE Hence the triangle is isoceles.AEG≅ACGTherefore from ASA theoremACG≅AEG

$A B D F$ is a square and $B C = E F$ in the given figure. Prove that
i) $Δ A B C ≅ Δ A F E$
ii) $Δ A C G ≅ Δ A E G$

Given,

$B C = E F$

Since ABCD is a square, sides are equal

$A F = A B$

$∠ B = ∠ F$

Therefore from SAS theorem

$A B C ≅ A F E$

$⟹ A C = A E$ Hence the triangle is isoceles.
$A E G ≅ A C G$

Therefore from ASA theorem

$A C G ≅ A E G$