ABC is an isosceles with AB= AC and D is a point on BC such that AD perpendicular BC prove that (i) BD = DC (ii) AD bisect angle BAD
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ABC is an isosceles with AB= AC and D is a point on BC such that AD perpendicular BC prove that (i) BD = DC (ii) AD bisect angle BAD
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Answer:
Given: ABC is an isosceles triangle with AB = AC and D is a point on BC such that BD = DC.
To prove that: AD bisects <A.
Proof: In ΔABD and ΔACD
AB = AC [given]
BD = DC [given]
AD is common.
So ΔABD and Δ ACD are congruent [By SSS rule]
Therefore <BAD = <CAD or AD bisects <A.
Answer:
Given:
To Find:
Solution:
AD||BC
∆ABD and ∆ACD are right angle triangle
Applying Pythagoras theorem
BD {}^{2} =AB {}^{2} -AD {}^{2} }
AB=AC
= > BD {}^{2} = AC { - }^{2} AD {}^{2}
AC {}^{2} AD {}^{2} = CD {}^{2}
= > BD {}^{2} = CD {}^{2}
= > BD = CD \:Eq..1) }
Compare∆ABD and ∆ACD
AB=AC (given)
BD=CD (From Eq..1)
AD=AD (Common)
∆ABD=∆ACD
∆BAD=∆CAD
QED
HENCE PROVED