Question

if the bisector of the vertical angle of a triangle bisects the base , prove that triangle is an isosceles​


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Answer 1

Answer:

That's your answer dude

Step-by-step explanation:

Given △ABC, AD is a bisector of ∠A which meets base BC at D such that BD=DC.

Produce AD to meet E such that AD=ED.

Now, in △ABD and △DEC

BD=DC ...... [Given]

AD=DE ........ [By construction]

∠ADB=∠EDC ..... [Vertically opposite angles]

∴ △ABD≅△EDC [∵SAS congruence ]

⟹AB=EC and ∠BAD=∠DEC ..... [CPCT]

Also, ∠BAD=∠DAC

⟹∠DAC=∠DEC

⟹ In △ACE, ∠AEC=∠CAE

⟹AC=CE ........ [Sides opposite to equal angles]

⟹AB=AC

Hence, △ABC is isosceles

Answer 2

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$\huge{\underline{\underline{\sf{\orange{</p><p>Question:-}}}}}$

if the bisector of the vertical angle of a triangle bisects the base , prove that triangle is an isosceles

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$\huge{\underline{\underline{\sf{\orange{Solution:-}}}}}$

${\blue{\sf\underline{Given}}}$

• A ∆ABC in which AD is the bisector of ∠A, which meets BC in D such that BD = DC

${\blue{\sf\underline{To\:prove}}}$

• AB = AC.

${\blue{\sf\underline{Construction}}}$

• Prove AD to E such that AD = DE .Join EC.

${\blue{\sf\underline{proof}}}$

In ∆ABD and ∆ECD, we have

$BD = DC \: \: \: \: (given)$

$AD = DE \: \: \: \: (by \: construction)$

$\angle \: ADB = \angle \: EDC \: \: \:$

$∴ \triangle ABD \cong \triangle ECD \: \: \: \: \: \: (SAS-criteria)$

$\therefore AB=EC and \angle 1=\angle 3 \: \: \: \: (c p.c.t.).$

Also,∠1 = ∠2. [∴ AD bisects ∠A]

∴ ∠2 = ∠3.

But,we know that the sides opposite to equal angles of a triangle are equal.

∴EC=AC.

So,AB=AC [∴EC=AB].

Hence,∆ABC is isosceles.

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