Question

in an isosceles triangle prove that the median are always greater than that of perpendicular.

MATHEMATICS

please help me

(sorry subject is taken SSC by mistake)

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Answer 1

Answer:

R.E.F image

Let AD be the median

of △ ABC

∴ In △ ABD & △ADC

AB=AC[isosceles triangle]

BD=DC [median]

∠ABD=∠ACD [ angle of isosceles triangle]

∴△ABD=△ADCso∠ADB=∠ADB

so∠ADB+∠ADC=180

∘

[liner angle]

so∠ADC=90

∘

[median is perpendicular]

Hence proved.

solution

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