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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Solution

In ΔABD and Δ ACD

AB = AC [given]

BD = CD [given]

AD = AD [common]

By SSS Congruence Criterion Rule

Δ ABD ≅ Δ ACD

∠ BAD = ∠CAD [CPCT]

∠ BAP = ∠CAP [CPCT] …

(ii)In ΔABP and Δ ACP

AB = AC [given]

∠ BAP = ∠CAP [proved above]

AP = AP [common]

By SAS Congruence Criterion Rule

Δ ABP ≅ Δ ACP

BP = CP [CPCT] … 2

∠APB = ∠APC [CPCT]

(iii) ∠ BAP = ∠CAP [From eq. 1]

Hence, AP bisects ∠ A.

Now, In Δ BDP and Δ CDP

BD = CD [given]

BP = CP [given]

DP = DP [common]

By SSS Congruence Criterion Rule

Δ BDP ≅ Δ CDP

∠ BDP = ∠CDP [CPCT]

AP bisects ∠ D.

(iv) AP stands on B

∠APB + ∠APC = 180degree

∠APB +∠APB = 180degree[proved above]

∠APB = 180degree /2

∠APB = 90degree

AP is the perpendicular bisector of BC

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Similar questions

△ 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

. If

A D

is extended to intersect

B C

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

$△ A B C$ 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 $B C$ .

ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that

(i) ΔABD ≅ ΔACD

(ii) ΔABP ≅ ΔACP

(iii) AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC.