in triangle ABC , angle C= 90 and AD , BE are medians through A and B respectively prove that 4AD^2=4AC^2+ BC^2

Question

in triangle ABC , angle C= 90 and AD , BE are medians through A and B respectively prove that 4AD^2=4AC^2+ BC^2

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Everleigh 1 month 2021-08-12T12:01:03+00:00 2 Answers 0 views 0

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    0
    2021-08-12T12:02:43+00:00

    Step-by-step explanation:

    ABC is a right angled triangle in which angle A=90° and AD is perpendicular

    to BC. Thus , In right angled triangle ABC , AB^2+AC^2=BC^2……………(1)

    In right angled triangle ADB

    AD^2+DB^2=AB^2……………….(2)

    In right angled triangle ADC

    AD^2+DC^2=AC^2……………….(3)

    by adding the eqn. (2) and (3)

    2.AD^2+DB^2+DC^2= AB^2+AC^2. , putting AB^2+AC^2=BC^2 from eqn. (1)

    2.AD^2+DB^2+DC^2=BC^2. , putting BC=BD+CD

    2.AD^2+DB^2+DC^2=(BD+CD)^2

    or. 2.AD^2+DB^2+DC^2=BD^2+CD^2+2×BD×CD.

    or. 2.AD^2 = 2×BD×CD.

    or. AD^2= BD×CD. Proved.

    Second -Method:-

    In right angled triangle ABC let angle ABC= B then angle ACB= 90°-B .

    Thus , in right angled triangle ADC , angle CAD = B.

    In right angled triangle ADB. , tanB=AD/BD………………(1)

    In right angled triangle CDA. , tanB=CD/AD………………..(2)

    From eqn. (1) and (2)

    AD/BD= CD/AD

    or. AD^2=BD×CD. Proved.

    0
    2021-08-12T12:03:01+00:00

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