   Chapter 5.3, Problem 35E Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Solutions

Chapter
Section Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

Use a two-column proof to prove the following theorem:“The lengths of the corresponding altitudes of similar triangles have the same ratio as lengths of any pair of corresponding sides.”Given:                           Δ D E F ∼ Δ M N P ; D G ¯ and M Q ¯ are altitudesProve:                             D G M Q = D E M N To determine

To prove:

The theorem “The lengths of the corresponding altitudes of similar triangles have the same ratio as lengths of any pair of corresponding sides.” By using two-column proof and prove DGMQ=DEMN by using given information.

Explanation

Definition:

AA:

If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

CSSTP:

Corresponding sides of similar triangles are proportional.

CASTC:

Corresponding angles of similar triangles are congruent.

Description:

Given that ΔDEFΔMNP, DG¯ and MQ¯ are altitudes.

The given figure is shown below.

Figure

From the given figure, it is observed that DG¯ and MQ¯ are altitudes.

It is known that an altitude is a line segment through a vertex and perpendicular to a line of base. That is, DG¯EF¯ and MQ¯NP¯.

Since, perpendicular lines form a right angle triangle, DGE and MQN are right angle triangles.

Right angle triangles are congruent. So that DGE=MQN.

From the above mentioned CASTC definition, EN since corresponding angles of similar triangles are congruent.

The above mentioned AA definition, the two triangles DGE and MQN are similar since the two angles of one triangle are congruent to two angles of another triangle. Hence, ΔDGEΔMQN.

From the definition of CSSTP, corresponding sides of similar triangles are proportional.

That is, DGMQ=DEMN

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