Question

In Fig.7.4, D and E are points on side BC of a Δ ABC such that BD = CE and AD = AE. 

Show that Δ ABD ≅ Δ ACE

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Answer 1

Answer:

BD=CE (given)

AD=AE (given)

angle A = angle A (common)

SAS condition apply

therefore ∆ ABD = ∆ACE

Answer 2

Given -

$BD = CE \: and \: AD = AE$

To show -

$\triangle \: ABD \: \cong \: \triangle \: ACE$

Solution -

$AD = AE$[Given]

$\angle \: ADE = \angle \: AED (Anngle \: opposite \: to \: equal \: sides \: are \: equal)....1)$

Now,

$\angle \: ADB + \angle \: ADE = 180(Linear \: pair)$

$\angle \: ADB = 180 - \angle \: ADE$

$\angle \: ADB = 180 - \angle \: AED(by \: 1)$

$In \: \triangle \: ABD \: and \: AEC$

$BD = CE \: (Given)$

$\angle \: ADB = \angle \: AEC (\therefore \: \angle \: AEC + \: \: \: \angle \: AED = 180 \: linear \: pair \: axiom)$

$AD = AE(Given)$

$\therefore \: \triangle \: ADB \: \cong \: \triangle \: AEC \: ( \: BY \: S. A. S\: Congruence)$

Additional information -

1) Axiom- Axiom are true statements that need no proof and are self evident.

For any two triangles to be congruent, the following four axioms are sufficient.

Axiom 1(SSS property) - If three sides of one triangle are equal respectively to the corresponding three sides of the other, then the triangles are congruent.

Axiom 2(SAS Property) - If two sides and the angle of one triangle are equal respectively to the corresponding two sides and the included angle of the other, then the two triangles are congruent,

Axiom 3(ASA property) - If two angles and included side of one triangle are equal to the angles and the included side of other triangle, then the two triangles are congruent.

Axiom 4(RHS property) - If the hypotenuse and one side of a right angled triangle are equal to the corresponding hypotenuse and one side of the other triangle, then the two right triangles are congruent.