The given figure shows two isosceles triangles ABC and DBC with common base BC. AD is extended to intersect BC at point P. Show that :
(i) △ABD ≡ △ACD.
(ii) △ABP ≡ △ACP.
(iii) AP bisects ∠BDC.
(iv) AP is perpendicular bisector of BC.

Question

The given figure shows two isosceles triangles ABC and DBC with common base BC. AD is extended to intersect BC at point P. Show that :
(i) △ABD ≡ △ACD.
(ii) △ABP ≡ △ACP.
(iii) AP bisects ∠BDC.
(iv) AP is perpendicular bisector of BC.

KnowledgeBoat · Accepted Answer

Given: ABC and DBC with common base BC. AD is extended to intersect BC at point P. (i) To prove: △ABD ≡ △ACD. Proof: In △ABD and △ACD, AB = AC (△ABC is an isosceles) AD = AD (Common side) BD = DC (△BDC is an isosceles) By SSS congruency criterion, Hence, △ABD ≅ △ACD. (ii) To prove: △ABP ≡ △ACP. Proof: From (i), △ABD ≅ △ACD By corresponding parts of congruent triangles, ∠BAD = ∠CAD i.e., ∠BAP = ∠PAC In △ABP and △ACP, AB = AC (△ABC is an isosceles) AP = AP (Common side) ∠BAP = ∠PAC (proved above) By SAS congruency criterion, Hence, △ABP ≅ △ACP. (iii) To prove: AP bisects ∠BDC. Proof: From (ii), △ABP ≅ △ACP By corresponding parts of congruent triangles, ⇒ BP = PC In △BDP and △CDP, DP = DP (common side) BP = PC (proved above) BD = DC (△BDC is an isosceles) By SSS congruency criterion, Hence, △BDP ≅ △CDP. By corresponding parts of congruent triangles, ∠BDP = ∠CDP Hence, AP bisects ∠BDC. (iv) To prove: AP is perpendicular bisector of BC. Proof: We know that ∠APB and ∠APC form linear pair. ⇒ ∠APB + ∠APC = 180° Also, ∠APB = ∠APC ∵ ∠APB = ∠APC = = 90° BP = PC and ∠APB = ∠APC = 90° Hence, AP is perpendicular bisector of BC.

Mathematics

The given figure shows two isosceles triangles ABC and DBC with common base BC. AD is extended to intersect BC at point P. Show that :
(i) △ABD ≡ △ACD.
(ii) △ABP ≡ △ACP.
(iii) AP bisects ∠BDC.
(iv) AP is perpendicular bisector of BC.

Triangles

Related Questions

In △ABC, AB = AC and D is a point in side BC such that AD bisects angle BAC.
Show that AD is perpendicular bisector of side BC.

In the given figure, BC = CE and ∠1 = ∠2.
Prove that : △GCB ≡ △DCE.

Two sides AB and BC and median AD of triangle ABC are respectively equal to sides PQ and QR and median PN of △PQR. Show that :
(i) △ABD ≡ △PQN.
(ii) △ABC ≡ △PQR.

The given figure shows PQ = PR and ∠Q = ∠R
Prove that: △PQS ≡ △PRT.

Mathematics

The given figure shows two isosceles triangles ABC and DBC with common base BC. AD is extended to intersect BC at point P. Show that :(i) △ABD ≡ △ACD.(ii) △ABP ≡ △ACP.(iii) AP bisects ∠BDC.(iv) AP is perpendicular bisector of BC.

Triangles

Related Questions

In △ABC, AB = AC and D is a point in side BC such that AD bisects angle BAC.Show that AD is perpendicular bisector of side BC.

In the given figure, BC = CE and ∠1 = ∠2.Prove that : △GCB ≡ △DCE.

Two sides AB and BC and median AD of triangle ABC are respectively equal to sides PQ and QR and median PN of △PQR. Show that :(i) △ABD ≡ △PQN.(ii) △ABC ≡ △PQR.

The given figure shows PQ = PR and ∠Q = ∠RProve that: △PQS ≡ △PRT.

The given figure shows two isosceles triangles ABC and DBC with common base BC. AD is extended to intersect BC at point P. Show that :
(i) △ABD ≡ △ACD.
(ii) △ABP ≡ △ACP.
(iii) AP bisects ∠BDC.
(iv) AP is perpendicular bisector of BC.

In △ABC, AB = AC and D is a point in side BC such that AD bisects angle BAC.
Show that AD is perpendicular bisector of side BC.

In the given figure, BC = CE and ∠1 = ∠2.
Prove that : △GCB ≡ △DCE.

Two sides AB and BC and median AD of triangle ABC are respectively equal to sides PQ and QR and median PN of △PQR. Show that :
(i) △ABD ≡ △PQN.
(ii) △ABC ≡ △PQR.

The given figure shows PQ = PR and ∠Q = ∠R
Prove that: △PQS ≡ △PRT.