Question

Que: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.

Answer 1

Answer:

Solution:

Given:

(i) BE and CF are altitudes.

(ii) AC = AB

To prove:

BE = CF

Proof:

Triangles ΔAEB and ΔAFC are similar by AAS congruency, since;

∠A = ∠A (common arm)

∠AEB = ∠AFC (both are right angles)

AB = AC (Given)

∴ ΔAEB ≅ ΔAFC

and BE = CF (by CPCT).

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Answer 2

Explanation:

Congruence of triangles:

Two ∆’s are congruent if sides and angles of a triangle are equal to the corresponding sides and angles of the other ∆.

 

In Congruent Triangles corresponding parts are always equal and we write it in short CPCT i e, corresponding parts of Congruent Triangles.

 

It is necessary to write a correspondence of vertices correctly for writing the congruence of triangles in symbolic form.

 

Criteria for congruence of triangles:

There are 4 criteria for congruence of triangles.

Here we use ASA congruence

ASA(angle side angle):

Two Triangles are congruent if two angles and the included side of One triangle are equal to two angles & the included side of the  other triangle.

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Given:

ΔABC is an isosceles∆ with AB = AC, BE and CF are altitudes.

To prove:

BE = CF

Proof:

In ΔAEB and ΔAFC,

∠A = ∠A (Common)

∠AEB = ∠AFC (each 90°)

AB = AC (Given)

Therefore, ΔAEB ≅ ΔAFC

(by AAS congruence rule)

Thus, BE = CF (by CPCT.)

 

Hence , altitudes BE & CF are equal.

 

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