Question

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​

Answer 1

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 2

Answer:

Given:

• ABC is an isosceles with AB= AC and D is a point on BC such that AD perpendicular BC

To Find:

• Show that ∆ BAD=∆CAD

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