In ∆ ABC and ∆ DEF, AB = DE and ∠A=∠D the two triangles will be congruent by SAS axiom if

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that:    

(i)    ∆AMC ≅ ∆BMD(ii)    ∠DBC is a right angle(iii)    ∆DBC ≅ ∆ACB

Given: In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B.

To Prove: (i) ∆AMC ≅ ∆BMD(ii)    ∠DBC is a right angle(iii)    ∠DBC ≅ ∆ACB

(i) In ∆AMC and ∆BMD,AM = BM| ∵ M is the mid-point of the hypotenuse ABCM = DM    | Given∠AMC = ∠BMD| Vertically Opposite Angles∴ ∆AMC ≅ ∆BMD. | SAS Rule(ii)    ∵ ∆AMC ≅ ∆BMD| From (i) above∠ACM = ∠BDM    | C.P.C.T.But these are alternate interior angles and they are equal∴ AC || BDNow, AC || BD and a transversal BC intersects them∴ ∠DBC + ∠ACB = 180°| ∵ The sum of the consecutive interior angles on the same side of a transversal is180°⇒ ∠DBC + 90° = 180°| ∵ ∠ACB = 90° (given)⇒     ∠DBC = 180° - 90° = 90°⇒ ∠DBC is a right angle.(iii)    In ∆DBC and ∆ACB,∠DBC = ∠ACB (each = 90°)| Proved in (ii) aboveBC = CB    | Common∵ ∆AMC ≅ ∆BMD    | Proved in (i) above∴ AC = BD    | C.P.C.T.∴ ∆DBC ≅ ∆ACB.    | SAS Rule(iv)    ∵ ∆DBC ≅ ∆ACB| Proved in (iii) above∴ DC = AB    | C.P.C.T.

      2CM = AB

In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom if ______.

• BC = EF

• AC = DE

• AC = EF

• BC = DE

In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom if AC = DE.

Explanation:

Given, in ΔABC and ΔDEF, AB = DF and ∠A = ∠D

We know that, two triangles will be congruent by ASA rule, if two angles and the included side of one triangle are equal to the two angles and the included side of other triangle.

∴ AC = DE

Concept: Congruence of Triangles

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