Question

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Answer 1

Answer:

Let ABC be an isosceles triangle such that

AB=

AC

.

Let AD be the bisector of ∠A.

To prove:-

BD=

DC

Proof:-

In △ABD&△ACD

AB= AC(∵△ABC is an isosceles triangle)

∠BAD= ∠CAD(∵AD is the bisector of ∠A)

AD= AD(Common)

By S.A.S.-

△ABD≅ △ACD

By corresponding parts of congruent triangles-

⇒ BD= DC

Hence proved that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.

Answer 2

➠ $\huge\underline\mathtt{Solution:-}$

Let ABC an isosceles triangle in which AB = AC .

Let AD be drawn perpendicular to BC .

it is required to prove that ∠BAD = ∠CAI

$\huge\underline\mathtt{proof:- }$

In an right angled triangles , A hypotenuse AB = hypotenuse AC .

AD is a common to both .

∴ ∆ABD ≅ ∆ACD

∴ ∠BAD = ∠ CAD and BD = CD.

✨

$\huge\underline\mathbb{Be Brainly }$ ✨