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Ne4ueva [31]
4 years ago
15

Complete the following proof related to the figure below.

Mathematics
2 answers:
Feliz [49]4 years ago
8 0

Answer:

HL

Step-by-step explanation:

We are given that

\angle A=\angle B=90^{\circ}

AC=BD

We have to find the postulate that is not used in line 3 reason in given proof.

In triangle MAC and BMD

Statement:\angle A=\angle B=90^{\circ}

AC=BD

Reason: Given

Statement:\angle AMC=\angle BMD

Reason: Vertical opposite angle theorem

Statement:\triangle AMC\cong \triangle BMD

Reason: AAS Postulate or LA postulate

AAS postulate: When two angles and one side of one triangle is equal to two corresponding angles and corresponding side of other triangle .Then , two triangles are congruent by AAS postulate.

LA postulate: When one acute angle and  one leg of one triangle is equal to its corresponding angle and corresponding leg of other triangle .Then, two triangles are congruent by LA postulate.

HL Postulate: When hypotenuse and one leg of one triangle is equal to its corresponding hypotenuse and corresponding leg of other triangle .Then, two triangles are congruent by HL postulate.

But we are not given MC=BM

Therefore, we cannot use HL postulate.

Statement:MC=BM

Reason:CPCT

Tju [1.3M]4 years ago
4 0

Answer:

HL

Step-by-step explanation:

Hypotenuse and leg of a right triangle (HL) :Two right triangles are called congruent if the hypotenuses and one corresponding leg are equal in both triangles.

Leg and angle of a right triangle ( LA ) : Two right triangles are called congruent if the corresponding angles and one corresponding leg are equal in both triangles,

AAS ( angle-angle-side) postulate of congruence : Two triangles are called congruent if two pairs of corresponding angles and a pair of opposite sides are equal in both triangles.

Here,

In triangles AMC and BMD,

∠A ≅ ∠B

∠AMC ≅ BMD

AC = BD

Thus, by the above definitions,

We can prove triangles AMC and BMD with help of both AAS and HL postulate,

But, we could not use HL postulate because MC = MD ( where MC and MD are the hypotenuse of triangles AMC and BMD respectively ) is not given.

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