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Standard X

Mathematics

Internal Angle Bisector Theorem

ABC and DBC...

Question

$△ A B C$ and $△ D B C$ are two isosceles triangles on the same base $B C$ and vertices $A$ and $D$ are on the same side of $B C$ as in figure.If $A D$ is extended to intersect $B C$ at $P,$ Show that $A P$ bisects $∠ A$ as well as $∠ D$

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Solution

Given: $△ A B C$ is isosceles,

$A B = A C$ ..... $(1)$

Also, $△ D B C$ is isosceles,

$D B = D C$ ..... $(2)$

Proof:In $△ A B D$ and $△ D B C$ , we have

$A B = A C$ from $(1)$

$B D = D C$ from $(2)$

$A D = A D$ (common)

$△ A B D ≅ △ A C D$ by SSS congruence rule.

So, $∠ B A P = ∠ P A C$ by CPCT ......... $(3)$

In $△ A B P$ and $△ A C P,$

$A B = A C$ (given)

$∠ B A P = ∠ P A C$ by CPCT (from $(3)$ )

$A P = A P$ (common)

$△ A B P ≅ △ A C P$ by SAS congruence rule.

$B P = C P$ ...... $(4)$

In $△ B D P$ and $△ C D P$ , we have

$B D = C D$ (Given)

$B P = C P$ (from $(4)$ )

$D P = D P$ (common)

So, $△ B D P ≅ △ C D P$ by SSS congruence rule

$\Rightarrow ∠ B D P = ∠ P D C$ by CPCT

Thus, $A P$ bisects $∠ D$

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P,

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are two isosceles triangles on the same base

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and vertices

A

and

D

are on the same side of

B C

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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 $B C$ .

ABC and DBC are two isosceles triangle on the same base BC and vertices A and D are on the same side of BC if AD is extended to intersect BC at P show that triangle ABD is congruent to triangle ACD and triangle ABD is congruent to triangle ACP and AP bisects angleA as well as angleD and AP is the perpendicular bisector of BC

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△ABC and △DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC as in figure.If AD is extended to intersect BC at P,Show that AP bisects ∠A as well as ∠D

$△ A B C$ and $△ D B C$ are two isosceles triangles on the same base $B C$ and vertices $A$ and $D$ are on the same side of $B C$ as in figure.If $A D$ is extended to intersect $B C$ at $P,$ Show that $A P$ bisects $∠ A$ as well as $∠ D$