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.
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Answer 1

Step-by-step explanation:

part 1

In ΔBCF andΔCBE,

∠BFC=∠CEB (Each 90º)

Hyp. BC= Hyp. BC (Common Side)

Side FC= Side EB (Given)

∴ By R.H.S. criterion of congruence, we have

ΔBCF≅ΔCBE

∴∠FBC=∠ECB (CPCT)

In ΔABC,

∠ABC=∠ACB

[∵∠FBC=∠ECB]

∴AB=AC (Converse of isosceles triangle theorem)

∴ ΔABC is an isosceles triangle.....

part 2

In △OMX and △ONX,

∠OMX=∠ONX=90

∘

OX=OX(common)

OM=ON where AB and CD are equal chords and equal chords are equidistant from the centre.

△OMX≅△ONX by RHS congruence rule.

∴∠OXM=∠OXN

i.e.,∠OXA=∠OXD

Hence proved....

hope it helps...