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In Delta ABC, AD is the perpendicular b...

In `Delta ABC`, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that `Delta ABC`is an isosceles triangle in which `A B\ =\ A C`.

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Given: `AD` is the perpendicular bisector of BC means `∠ADB = ∠ADC = 90^@` and `BD = DC`
To Prove: `ΔABC` is an isosceles triangle in which `AB = AC`.
In `ΔADC` and `ΔADB`, ...
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Knowledge Check

  • In the adjoning figure of DeltaABC, AD is the perpendicular bisector of side BC. The triangle ABC is :

    A
    right angled
    B
    isosceles
    C
    scalene
    D
    equilateral

  • In triangle ABC, AD is prependicular to BC and DE is perpendicular to AB

    A
    `{:(a,b,c,d),(p,r,q,q):}`
    B
    `{:(a,b,c,d),(q,r,p,s):}`
    C
    `{:(a,b,c,d),(s,p,q,r):}`
    D
    `{:(a,b,c,d),(r,p,s,q):}`

  • In a Delta ABC , the median AD is perpendicular to AC. If b = 5 and c = 11, then a =

    A
    10
    B
    12
    C
    14
    D
    `sqrt(221)`

    • Similar Questions

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    AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that AD bisects BC

    In Delta ABC , if (cos A)/(a) = (cos B)/(b) , then show that it is an isosceles triangle.

    D is a point on side BC of Delta ABC such that AD=AC (see Fig.7.47). Show that AB>AD

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