$Δ A B C$ and $Δ D B C$ are two isosceles triangle on the same base $B C$ and vertices $A$ and $D$ are on the same side of $B C$ (see fig $7.39$ ). If $A D$ is extended to intersect $B C$ at $P$ , show that
(i) $Δ A B D ≅ Δ A C D$
(ii) $Δ A B P ≅ Δ A C P$
(ii) $A P$ bisects $∠ A$ as wll as $∠ D$ .
(Iv) $A P$ is the perpendicular bisector of $B C$ .

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Solution

$(i)$ In $△ A B D$ and $△ A C D,$
$A D = A D$ [ Common side ]
$A B = A C$ [ Given ]
$B D = C D$ [ Given ]
$∴$ $△ A B D ≅ △ A C D$ [ Bu SSS congruence rule ]
$∠ B A D = ∠ C A D$ [ CPCT ] ------ ( 1 )
$i . e . ∠ B A P = ∠ C A P$ --------- ( 2 )

$(i i)$ In $△ A B P$ and $△ A C P$ ,
$A P = A P$ [ Common side ]
$∠ B A P = ∠ C A P$ [ From ( 2 ) ]
$A B = A C$ [ Given ]
$∴$ $△ A B P ≅ A C P$ [ By SAS Congruence rule ]
$B P = C P$ [ By CPCT ] --------- ( 3 )

$(i i i)$ $∠ B A D = ∠ C A D$ [ From ( 1 )]
$∴$ $A P$ bisects $∠ A$ .
In $△ B P D$ and $△ C P D$ ,
$P D = P D$ [ Common side ]
$B D = C D$ [ Given ]
$B P = C P$ [ From ( 3 ) ]
$∴$ $△ B P D ≅ △ C P D$ [ By SSS Congruence rule ]
$∠ B D P = ∠ C D P$ [ By CPCT ] ------- ( 4 )

$(i v)$ $∠ B P D = ∠ C P D$ [ From ( 4 ) ]
$B P = C P$ [ From ( 3 ) ]
$∠ B P D + ∠ C P D = 180^{o}$
$2 ∠ B P D = 180^{o}$
$∴$ $∠ B P D = 90^{o}$
$∴$ $A P$ is the perpendicular bisector of $B C$ .

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Similar questions

Δ A B C

and

Δ D B C

are two isosceles

I

triangles on the same base

B C

and vertices

A

and

D

are on the same side of

B C

(see figure). If

A D

is extended to intersect

B C

P

, show that
(i)

Δ A B D ≅ Δ A C D

(ii)

Δ A B P ≅ Δ A C P

(iii)

A P

bisects

∠ A

as well as

∠ D

(iv)

A P

is the perpendicular bisector of

B C

$△ A B C$ and $△ D B C$ are two isosceles triangles on the same base $B C$ (see in fig.) If $A D$ is extended to intersect $B C$ at $P$ , show that
(i) $△ A B D ≅ △ A C D$
(ii) $△ A B P ≅ △ A C P$
(iii) $A P$ bisects $∠ A$ as well as $∠ D$ .
(iv) $A P$ is the perpendicular bisector of $B C$ .