Question

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles. OR If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.​

Answer 1

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Consider triangle BFC AND CEB

BC = CB. ( COMMON )

ANGLE BFC = ANGLE CEB = 90 (GIVEN)

CF = BE. ( GIVEN )

SO, triangle BFC is congruent to triangle CEB. ( RHS )

Hence , BF = CE. ( CPCT )_____________(1)

and, considered triangle ABE and ACF

angle BAC = angle CAB. ( COMMON )

angle AFC = angle AEB. (GIVEN)

CF = BE. (GIVEN)

HENCE,. triangle ABE is congruent to triangle ACF. (AAS)

So, AF = AE. (CPCT). _______________(2)

adding (1) and (2)

we get,

BF + AF = CE + AE

AB = AC

hence , triangle ABC is iso. triangle.