Question

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal show that 1 triangle ABE IS unequal to triangle ACF

Answer 1

Step-by-step 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 in which BE perpendicular to AC & CF perpendicular to AB, such that BE=CF.

To Prove:

i) ΔABE ≅ ΔACF

ii) AB=AC

Proof:

(i) In ΔABE & ΔACF,

∠A = ∠A (Common)

∠AEB = ∠AFC (each 90°)

BE = CF (Given)

Therefore, ΔABE ≅ ΔACF (by ASA congruence rule)

(ii) since ΔABE ≅ ΔACF

Thus, AB = AC (by CPCT)

Therefore ∆ABC is an isosceles triangle.

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