$Δ A B C$ 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 $B C$ at $P$ , 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 $B C$ .

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Solution

(i) In $Δ^{'} s$ $A B D$ and $A C D$ , we have
$A B = A C$ (given)
$B D = D C$ (given)
and $A D = A D$ (common)
$∴ Δ A B D ≅ Δ A C D$
(by SSS congruency rule)
(ii) In $Δ^{'} s$ $A B P$ and $A C P$ , we have
$A B = A C$ (given)
$∠ B A P = ∠ C A P$
and $A P = A P$ (common)
$[∵ Δ A B D ≅ Δ A C D, ∠ B A D = ∠ C A D, ∠ B A P = ∠ C A P]$
$∴ Δ A B P ≅ Δ A C P$
(by SAS congruency rule)
(iii) We have already proved in (i) that
$Δ A B D ≅ Δ C A D$
$∠ B A P = ∠ C A P$
$A P$ bisects $∠ A$ i.e. $A P$ is the bisector of $∠ A$ .
In $Δ^{'} s$ $B D P$ and $C D P$ , we have
$B D = C D$ (given)
$B P = C P [∵ Δ A B P = Δ A C P]$
and $D P = D P$ (common)
$∴ Δ B D P ≅ Δ C D P$
(by SSS congruency rule)
$∠ B D P = ∠ C D P$
$D P$ is the bisector of $∠ D$ .
Hence, $A P$ is the bisector of $∠ A$ as well as $∠ D$ .
(iv) In (iii), we have proved that
$Δ B D P ≅ Δ C D P$
$B P = C P$ and $∠ B P D = ∠ C P D = 90^{0}$ .
$∴ ∠ B P D$ and $∠ C P D$ form a linear pair
$D P$ is the perpendicular bisector of $B C$
Hence, $A P$ is the perpendicular bisector of $B C$

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Δ A B C

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P

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Q. Question 1 (iii)

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$△ A B C$ 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 $B C$ .