In $Δ A B C, A D$ is the perpendicular bisector of $B C$ . Show that $Δ A B C$ is an isosceles triangle in which $A B = A C$

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Solution

In, $△ A D B$ and $△ A D C$ ,

$B D = D C$ (Given)

$A D = A D$ (Common)

$A D B = A D C$ (Given)

$∴ △ A B D ≅ △ A D C (S A S c o n g w e n i y)$

$∴ A B = A C (C . P . C . T)$

Hence $△ A B C$ is isosceles

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Similar questions

In ΔABC, AD is the perpendicular bisector of BC (see the given figure). Show that ΔABC is an isosceles triangle in which AB = AC.

Question 2
In $Δ A B C, A D$ is the perpendicular bisector of BC (see the given figure).
Show that $Δ A B C$ is an isosceles triangle in which AB = AC.

Q. In

Δ A B C, A D

is the perpendicular bisects of BC sent Fig. show that

Δ A B C

is an isosceles triangle in which AB = AC.

Q. In

Δ

ABC, AD is the perpendicular bisector of BC(see adjacent figure). Show that

Δ

ABC is an isosceles triangle in which AB

=

AC.

Q. In

△ A B C, A D

is the perpendicular bisector of

B C

. Show that

△ A B C

is an isosceles triangle in which

A B = A C

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In ΔABC,AD is the perpendicular bisector of BC. Show that ΔABC is an isosceles triangle in which AB=AC

In, △ADB and △ADC, BD=DC (Given) AD=AD (Common) ADB=ADC (Given) ∴△ABD≅△ADC(SAScongweniy) ∴AB=AC(C.P.C.T) Hence △ABC is isosceles

In $Δ A B C, A D$ is the perpendicular bisector of $B C$ . Show that $Δ A B C$ is an isosceles triangle in which $A B = A C$

In, $△ A D B$ and $△ A D C$ ,

$B D = D C$ (Given)

$A D = A D$ (Common)

$A D B = A D C$ (Given)

$∴ △ A B D ≅ △ A D C (S A S c o n g w e n i y)$

$∴ A B = A C (C . P . C . T)$

Hence $△ A B C$ is isosceles