Question

ABC is an isosceles triangle in which altitudes BD and CE
are drawn to equal sides AC and AB respectively (see figure)
Show that these altitudes are equal.​

Answer 1

Explanation:

GIVEN:

ABC IS AN ISOSCELES

AB=AC

BE AND CF ARE ALTITUDES

SO, angle AEB and angle AFC ARE EQUAL TO 90°

TO PROVE:

BE=CF

PROOF:

WE TAKE ∆ABE AND ∆ACF,

IN ∆ABE AND ∆ACF,

angleAEB=AFC

angke A=angleA

AB=AC

therefore,

∆ABE=~∆ACF. (AAS COGURENCE....)

BE=CF (BY CPCT)

HENCE,PROVED

Answer 2

Explanation:

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