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
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
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
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
