If D is the mid point of side BC of a triangle ABC and AD is the
perpendicular to AC then:

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(A) 3AC

^{2}= BC^{2}-AB^{2}
(B) 3BC

^{2}= AC^{2}-3AB^{2}
(C) BC

(D) None of the above^{2}+ AC^{2}= 5AB^{2}To enter your answer, click on "comments" below:

A

ReplyDeleteD

ReplyDeletesee consider a triangle with base 8cm(BC) and altitude 3(let be AD).

since D bisects and form right angle,we have 2 RIGHT ANGLED TRIANGLE with height AD(=3cm)

bases BD(=4cm) and DC(=4cm).

then using pythogorus theorem,we get AB=5cm and AD=5cm

subtituting the values in the above four options ,i am getting none of them satisfied

SO answer option is D)NONE of the above

In the question, it is given that AD is perpendicular to AC but not to BC...

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